diff --git a/project_euler/problems/025_1000_digit_fibonacci_number.txt b/project_euler/problems/025_1000_digit_fibonacci_number.txt index 9c19735..c009c88 100644 --- a/project_euler/problems/025_1000_digit_fibonacci_number.txt +++ b/project_euler/problems/025_1000_digit_fibonacci_number.txt @@ -5,17 +5,17 @@ http://projecteuler.net/problem=025 The Fibonacci sequence is defined by the recurrence relation: Fn = Fn1 + Fn2, where F1 = 1 and F2 = 1. Hence the first 12 terms will be: -F1 = 1 -F2 = 1 -F3 = 2 -F4 = 3 -F5 = 5 -F6 = 8 -F7 = 13 -F8 = 21 -F9 = 34 -F10 = 55 -F11 = 89 +F1 = 1 +F2 = 1 +F3 = 2 +F4 = 3 +F5 = 5 +F6 = 8 +F7 = 13 +F8 = 21 +F9 = 34 +F10 = 55 +F11 = 89 F12 = 144 The 12th term, F12, is the first term to contain three digits. What is the first term in the Fibonacci sequence to contain 1000 digits? diff --git a/project_euler/problems/057_square_root_convergents.txt b/project_euler/problems/057_square_root_convergents.txt index db3ea10..ebf79ab 100644 --- a/project_euler/problems/057_square_root_convergents.txt +++ b/project_euler/problems/057_square_root_convergents.txt @@ -5,9 +5,9 @@ Square root convergents It is possible to show that the square root of two can be expressed as an infinite continued fraction. 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... By expanding this for the first four iterations, we get: -1 + 1/2 = 3/2 = 1.5 -1 + 1/(2 + 1/2) = 7/5 = 1.4 -1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666... +1 + 1/2 = 3/2 = 1.5 +1 + 1/(2 + 1/2) = 7/5 = 1.4 +1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666... 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379... The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator. In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator? diff --git a/project_euler/problems/066_diophantine_equation.txt b/project_euler/problems/066_diophantine_equation.txt index 5d14b9b..962cf08 100644 --- a/project_euler/problems/066_diophantine_equation.txt +++ b/project_euler/problems/066_diophantine_equation.txt @@ -7,10 +7,10 @@ x2 – Dy2 = 1 For example, when D=13, the minimal solution in x is 6492 – 131802 = 1. It can be assumed that there are no solutions in positive integers when D is square. By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following: -32 – 222 = 1 +32 – 222 = 1 22 – 312 = 1 -92 – 542 = 1 -52 – 622 = 1 +92 – 542 = 1 +52 – 622 = 1 82 – 732 = 1 Hence, by considering minimal solutions in x for D 7, the largest x is obtained when D=5. Find the value of D 1000 in minimal solutions of x for which the largest value of x is obtained. diff --git a/project_euler/problems/084_monopoly_odds.txt b/project_euler/problems/084_monopoly_odds.txt index f7a0413..e368a4e 100644 --- a/project_euler/problems/084_monopoly_odds.txt +++ b/project_euler/problems/084_monopoly_odds.txt @@ -82,13 +82,13 @@ A player starts on the GO square and adds the scores on two 6-sided dice to dete In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail. At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square. -Community Chest (2/16 cards): +Community Chest (2/16 cards): Advance to GO Go to JAIL -Chance (10/16 cards): +Chance (10/16 cards): Advance to GO Go to JAIL diff --git a/project_euler/problems/090_cube_digit_pairs.txt b/project_euler/problems/090_cube_digit_pairs.txt index fb646b0..2f9a795 100644 --- a/project_euler/problems/090_cube_digit_pairs.txt +++ b/project_euler/problems/090_cube_digit_pairs.txt @@ -11,7 +11,7 @@ In fact, by carefully choosing the digits on both cubes it is possible to displa For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube. However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09. In determining a distinct arrangement we are interested in the digits on each cube, not the order. -{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5} +{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5} {1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9} But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers. How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed? diff --git a/project_euler/problems/141_investigating_progressive_numbers_n_which_are_also_square.txt b/project_euler/problems/141_investigating_progressive_numbers_n_which_are_also_square.txt index 3e631e4..d74dedb 100644 --- a/project_euler/problems/141_investigating_progressive_numbers_n_which_are_also_square.txt +++ b/project_euler/problems/141_investigating_progressive_numbers_n_which_are_also_square.txt @@ -3,7 +3,7 @@ http://projecteuler.net/problem=141 Investigating progressive numbers, n, which are also square A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order. -For example, 58 divided by 6 has quotient 9 and remainder 4. It can also be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common ratio 3/2). +For example, 58 divided by 6 has quotient 9 and remainder 4. It can also be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common ratio 3/2). We will call such numbers, n, progressive. Some progressive numbers, such as 9 and 10404 = 1022, happen to also be perfect squares. The sum of all progressive perfect squares below one hundred thousand is 124657. Find the sum of all progressive perfect squares below one trillion (1012). diff --git a/project_euler/problems/147_rectangles_in_crosshatched_grids.txt b/project_euler/problems/147_rectangles_in_crosshatched_grids.txt index a76b414..92a2901 100644 --- a/project_euler/problems/147_rectangles_in_crosshatched_grids.txt +++ b/project_euler/problems/147_rectangles_in_crosshatched_grids.txt @@ -5,10 +5,10 @@ Rectangles in cross-hatched grids In a 3x2 cross-hatched grid, a total of 37 different rectangles could be situated within that grid as indicated in the sketch. There are 5 grids smaller than 3x2, vertical and horizontal dimensions being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids: -1x1: 1 -2x1: 4 -3x1: 8 -1x2: 4 +1x1: 1 +2x1: 4 +3x1: 8 +1x2: 4 2x2: 18 Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles could be situated within 3x2 and smaller grids. How many different rectangles could be situated within 47x43 and smaller grids? diff --git a/project_euler/problems/149_searching_for_a_maximum_sum_subsequence.txt b/project_euler/problems/149_searching_for_a_maximum_sum_subsequence.txt index 40ada27..35e4dd5 100644 --- a/project_euler/problems/149_searching_for_a_maximum_sum_subsequence.txt +++ b/project_euler/problems/149_searching_for_a_maximum_sum_subsequence.txt @@ -11,7 +11,7 @@ Looking at the table below, it is easy to verify that the maximum possible sum o Now, let us repeat the search, but on a much larger scale: First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator": -For 1 k 55, sk = [100003 200003k + 300007k3] (modulo 1000000) 500000. +For 1 k 55, sk = [100003 200003k + 300007k3] (modulo 1000000) 500000. For 56 k 4000000, sk = [sk24 + sk55 + 1000000] (modulo 1000000) 500000. Thus, s10 = 393027 and s100 = 86613. The terms of s are then arranged in a 20002000 table, using the first 2000 numbers to fill the first row (sequentially), the next 2000 numbers to fill the second row, and so on. diff --git a/project_euler/problems/150_searching_a_triangular_array_for_a_sub-triangle_having_minimum-sum.txt b/project_euler/problems/150_searching_a_triangular_array_for_a_sub-triangle_having_minimum-sum.txt index 55855a4..5ff3fe5 100644 --- a/project_euler/problems/150_searching_a_triangular_array_for_a_sub-triangle_having_minimum-sum.txt +++ b/project_euler/problems/150_searching_a_triangular_array_for_a_sub-triangle_having_minimum-sum.txt @@ -8,29 +8,29 @@ In the example below, it can be easily verified that the marked triangle satisfi We wish to make such a triangular array with one thousand rows, so we generate 500500 pseudo-random numbers sk in the range 219, using a type of random number generator (known as a Linear Congruential Generator) as follows: -t := 0 - -for k = 1 up to k = 500500: - +t := 0 + +for k = 1 up to k = 500500: +     t := (615949*t + 797807) modulo 220 - +     sk := t219 Thus: s1 = 273519, s2 = 153582, s3 = 450905 etc Our triangular array is then formed using the pseudo-random numbers thus: - + s1 - + s2  s3 - -s4  s5  s6  - + +s4  s5  s6 + s7  s8  s9  s10 - -... -Sub-triangles can start at any element of the array and extend down as far as we like (taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on). - -The "sum of a sub-triangle" is defined as the sum of all the elements it contains. - +... + +Sub-triangles can start at any element of the array and extend down as far as we like (taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on). + +The "sum of a sub-triangle" is defined as the sum of all the elements it contains. + Find the smallest possible sub-triangle sum. diff --git a/project_euler/problems/153_investigating_gaussian_integers.txt b/project_euler/problems/153_investigating_gaussian_integers.txt index 2cb9ac3..fbc0b92 100644 --- a/project_euler/problems/153_investigating_gaussian_integers.txt +++ b/project_euler/problems/153_investigating_gaussian_integers.txt @@ -2,45 +2,44 @@ http://projecteuler.net/problem=153 Investigating Gaussian Integers -As we all know the equation x2=-1 has no solutions for real x. - -If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i. - -If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i. - -x=3+2i and x=3-2i are called each others' complex conjugate. - -Numbers of the form a+bi are called complex numbers. - +As we all know the equation x2=-1 has no solutions for real x. + +If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i. + +If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i. + +x=3+2i and x=3-2i are called each others' complex conjugate. + +Numbers of the form a+bi are called complex numbers. + In general a+bi and abi are each other's complex conjugate. -A Gaussian Integer is a complex number a+bi such that both a and b are integers. - -The regular integers are also Gaussian integers (with b=0). - -To distinguish them from Gaussian integers with b 0 we call such integers "rational integers." - -A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer. - -If for example we divide 5 by 1+2i we can simplify in the following manner: - -Multiply numerator and denominator by the complex conjugate of 1+2i: 12i. - -The result is -. - -So 1+2i is a divisor of 5. - -Note that 1+i is not a divisor of 5 because . - +A Gaussian Integer is a complex number a+bi such that both a and b are integers. + +The regular integers are also Gaussian integers (with b=0). + +To distinguish them from Gaussian integers with b 0 we call such integers "rational integers." + +A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer. + +If for example we divide 5 by 1+2i we can simplify in the following manner: + +Multiply numerator and denominator by the complex conjugate of 1+2i: 12i. + +The result is . + +So 1+2i is a divisor of 5. + +Note that 1+i is not a divisor of 5 because . + Note also that if the Gaussian Integer (a+bi) is a divisor of a rational integer n, then its complex conjugate (abi) is also a divisor of n. -In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 2i, 2 + i, 2 i, 5}. - +In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 2i, 2 + i, 2 i, 5}. + The following is a table of all of the divisors for the first five positive rational integers: -n Gaussian integer divisors -with positive real partSum s(n) of these - +n Gaussian integer divisors +with positive real partSum s(n) of these + divisors 111 diff --git a/project_euler/problems/154_exploring_pascals_pyramid.txt b/project_euler/problems/154_exploring_pascals_pyramid.txt index bfac3a1..515b794 100644 --- a/project_euler/problems/154_exploring_pascals_pyramid.txt +++ b/project_euler/problems/154_exploring_pascals_pyramid.txt @@ -7,7 +7,7 @@ A triangular pyramid is constructed using spherical balls so that each ball rest Then, we calculate the number of paths leading from the apex to each position: A path starts at the apex and progresses downwards to any of the three spheres directly below the current position. Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it). -The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion +The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion (x + y + z)n. How many coefficients in the expansion of (x + y + z)200000 are multiples of 1012? diff --git a/project_euler/problems/155_counting_capacitor_circuits.txt b/project_euler/problems/155_counting_capacitor_circuits.txt index 76b03fe..3604b46 100644 --- a/project_euler/problems/155_counting_capacitor_circuits.txt +++ b/project_euler/problems/155_counting_capacitor_circuits.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=155 Counting Capacitor Circuits -An electric circuit uses exclusively identical capacitors of the same value C. - +An electric circuit uses exclusively identical capacitors of the same value C. + The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit. Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to n=3 capacitors of 60 F each, we can obtain the following 7 distinct total capacitance values: If we denote by D(n) the number of distinct total capacitance values we can obtain when using up to n equal-valued capacitors and the simple procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ... Find D(18). -Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +..., - -whereas when connecting them in series, the overall capacitance is given by: +Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +..., + +whereas when connecting them in series, the overall capacitance is given by: diff --git a/project_euler/problems/156_counting_digits.txt b/project_euler/problems/156_counting_digits.txt index 953f7e3..1230fc4 100644 --- a/project_euler/problems/156_counting_digits.txt +++ b/project_euler/problems/156_counting_digits.txt @@ -2,9 +2,9 @@ http://projecteuler.net/problem=156 Counting Digits -Starting from zero the natural numbers are written down in base 10 like this: - -0 1 2 3 4 5 6 7 8 9 10 11 12.... +Starting from zero the natural numbers are written down in base 10 like this: + +0 1 2 3 4 5 6 7 8 9 10 11 12.... Consider the digit d=1. After we write down each number n, we will update the number of ones that have occurred and call this number f(n,1). The first values for f(n,1), then, are as follows: @@ -38,16 +38,16 @@ nf(n,1) 125 -Note that f(n,1) never equals 3. - +Note that f(n,1) never equals 3. + So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The next solution is n=199981. -In the same manner the function f(n,d) gives the total number of digits d that have been written down after the number n has been written. - +In the same manner the function f(n,d) gives the total number of digits d that have been written down after the number n has been written. + In fact, for every digit d 0, 0 is the first solution of the equation f(n,d)=n. -Let s(d) be the sum of all the solutions for which f(n,d)=n. - +Let s(d) be the sum of all the solutions for which f(n,d)=n. + You are given that s(1)=22786974071. Find s(d) for 1 d 9. -Note: if, for some n, f(n,d)=n +Note: if, for some n, f(n,d)=n for more than one value of d this value of n is counted again for every value of d for which f(n,d)=n. diff --git a/project_euler/problems/157_solving_the_diophantine_equation_1a1b_p10n.txt b/project_euler/problems/157_solving_the_diophantine_equation_1a1b_p10n.txt index 3002b68..72f230b 100644 --- a/project_euler/problems/157_solving_the_diophantine_equation_1a1b_p10n.txt +++ b/project_euler/problems/157_solving_the_diophantine_equation_1a1b_p10n.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=157 Solving the diophantine equation 1/a+1/b= p/10n -Consider the diophantine equation 1/a+1/b= p/10n with a, b, p, n positive integers and a b. +Consider the diophantine equation 1/a+1/b= p/10n with a, b, p, n positive integers and a b. For n=1 this equation has 20 solutions that are listed below: 1/1+1/1=20/10 diff --git a/project_euler/problems/158_exploring_strings_for_which_only_one_character_comes_lexicographically_after_its_neighbour_to_the_left.txt b/project_euler/problems/158_exploring_strings_for_which_only_one_character_comes_lexicographically_after_its_neighbour_to_the_left.txt index f06d709..27dbef1 100644 --- a/project_euler/problems/158_exploring_strings_for_which_only_one_character_comes_lexicographically_after_its_neighbour_to_the_left.txt +++ b/project_euler/problems/158_exploring_strings_for_which_only_one_character_comes_lexicographically_after_its_neighbour_to_the_left.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=158 Exploring strings for which only one character comes lexicographically after its neighbour to the left -Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed. -Examples are 'abc', 'hat' and 'zyx'. -When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left. -For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left. For 'zyx' there are zero characters that come lexicographically after its neighbour to the left. +Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed. +Examples are 'abc', 'hat' and 'zyx'. +When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left. +For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left. For 'zyx' there are zero characters that come lexicographically after its neighbour to the left. In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left. -We now consider strings of n 26 different characters from the alphabet. +We now consider strings of n 26 different characters from the alphabet. For every n, p(n) is the number of strings of length n for which exactly one character comes lexicographically after its neighbour to the left. What is the maximum value of p(n)? diff --git a/project_euler/problems/159_digital_root_sums_of_factorisations.txt b/project_euler/problems/159_digital_root_sums_of_factorisations.txt index ac83ff9..4c5e634 100644 --- a/project_euler/problems/159_digital_root_sums_of_factorisations.txt +++ b/project_euler/problems/159_digital_root_sums_of_factorisations.txt @@ -1,23 +1,23 @@ http://projecteuler.net/problem=159 -Digital root sums of factorisations +Digital root sums of factorisations -A composite number can be factored many different ways. +A composite number can be factored many different ways. For instance, not including multiplication by one, 24 can be factored in 7 distinct ways: - -24 = 2x2x2x3 -24 = 2x3x4 -24 = 2x2x6 -24 = 4x6 -24 = 3x8 -24 = 2x12 -24 = 24 - -Recall that the digital root of a number, in base 10, is found by adding together the digits of that number, -and repeating that process until a number is arrived at that is less than 10. + +24 = 2x2x2x3 +24 = 2x3x4 +24 = 2x2x6 +24 = 4x6 +24 = 3x8 +24 = 2x12 +24 = 24 + +Recall that the digital root of a number, in base 10, is found by adding together the digits of that number, +and repeating that process until a number is arrived at that is less than 10. Thus the digital root of 467 is 8. -We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number. +We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number. The chart below demonstrates all of the DRS values for 24. FactorisationDigital Root Sum @@ -36,7 +36,7 @@ FactorisationDigital Root Sum 24 6 -The maximum Digital Root Sum of 24 is 11. -The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11. +The maximum Digital Root Sum of 24 is 11. +The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11. Find mdrs(n) for 1 n 1,000,000. diff --git a/project_euler/problems/160_factorial_trailing_digits.txt b/project_euler/problems/160_factorial_trailing_digits.txt index a7c0b00..93b9c9f 100644 --- a/project_euler/problems/160_factorial_trailing_digits.txt +++ b/project_euler/problems/160_factorial_trailing_digits.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=160 Factorial trailing digits -For any N, let f(N) be the last five digits before the trailing zeroes in N!. +For any N, let f(N) be the last five digits before the trailing zeroes in N!. For example, -9! = 362880 so f(9)=36288 -10! = 3628800 so f(10)=36288 +9! = 362880 so f(9)=36288 +10! = 3628800 so f(10)=36288 20! = 2432902008176640000 so f(20)=17664 Find f(1,000,000,000,000) diff --git a/project_euler/problems/161_triominoes.txt b/project_euler/problems/161_triominoes.txt index b4f502f..8fd311d 100644 --- a/project_euler/problems/161_triominoes.txt +++ b/project_euler/problems/161_triominoes.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=161 Triominoes -A triomino is a shape consisting of three squares joined via the edges. +A triomino is a shape consisting of three squares joined via the edges. There are two basic forms: If all possible orientations are taken into account there are six: -Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes. +Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes. If we consider tilings that can be obtained by reflection or rotation from another tiling as different there are 41 ways a 2 by 9 grid can be tiled with triominoes: In how many ways can a 9 by 12 grid be tiled in this way by triominoes? diff --git a/project_euler/problems/162_hexadecimal_numbers.txt b/project_euler/problems/162_hexadecimal_numbers.txt index f7ff895..48accb2 100644 --- a/project_euler/problems/162_hexadecimal_numbers.txt +++ b/project_euler/problems/162_hexadecimal_numbers.txt @@ -5,9 +5,9 @@ Hexadecimal numbers In the hexadecimal number system numbers are represented using 16 different digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F The hexadecimal number AF when written in the decimal number system equals 10x16+15=175. -In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present. +In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present. Like numbers written in base ten we write hexadecimal numbers without leading zeroes. -How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once? +How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once? Give your answer as a hexadecimal number. (A,B,C,D,E and F in upper case, without any leading or trailing code that marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F) diff --git a/project_euler/problems/163_crosshatched_triangles.txt b/project_euler/problems/163_crosshatched_triangles.txt index 443ea73..ff98e95 100644 --- a/project_euler/problems/163_crosshatched_triangles.txt +++ b/project_euler/problems/163_crosshatched_triangles.txt @@ -7,7 +7,7 @@ Consider an equilateral triangle in which straight lines are drawn from each ver Sixteen triangles of either different shape or size or orientation or location can now be observed in that triangle. Using size 1 triangles as building blocks, larger triangles can be formed, such as the size 2 triangle in the above sketch. One-hundred and four triangles of either different shape or size or orientation or location can now be observed in that size 2 triangle. It can be observed that the size 2 triangle contains 4 size 1 triangle building blocks. A size 3 triangle would contain 9 size 1 triangle building blocks and a size n triangle would thus contain n2 size 1 triangle building blocks. If we denote T(n) as the number of triangles present in a triangle of size n, then -T(1) = 16 +T(1) = 16 T(2) = 104 Find T(36). diff --git a/project_euler/problems/165_intersections.txt b/project_euler/problems/165_intersections.txt index 40ef49b..2b36685 100644 --- a/project_euler/problems/165_intersections.txt +++ b/project_euler/problems/165_intersections.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=165 Intersections -A segment is uniquely defined by its two endpoints. By considering two line segments in plane geometry there are three possibilities: +A segment is uniquely defined by its two endpoints. By considering two line segments in plane geometry there are three possibilities: the segments have zero points, one point, or infinitely many points in common. -Moreover when two segments have exactly one point in common it might be the case that that common point is an endpoint of either one of the segments or of both. If a common point of two segments is not an endpoint of either of the segments it is an interior point of both segments. -We will call a common point T of two segments L1 and L2 a true intersection point of L1 and L2 if T is the only common point of L1 and L2 and T is an interior point of both segments. +Moreover when two segments have exactly one point in common it might be the case that that common point is an endpoint of either one of the segments or of both. If a common point of two segments is not an endpoint of either of the segments it is an interior point of both segments. +We will call a common point T of two segments L1 and L2 a true intersection point of L1 and L2 if T is the only common point of L1 and L2 and T is an interior point of both segments. Consider the three segments L1, L2, and L3: -L1: (27, 44) to (12, 32) -L2: (46, 53) to (17, 62) +L1: (27, 44) to (12, 32) +L2: (46, 53) to (17, 62) L3: (46, 70) to (22, 40) It can be verified that line segments L2 and L3 have a true intersection point. We note that as the one of the end points of L3: (22,40) lies on L1 this is not considered to be a true point of intersection. L1 and L2 have no common point. So among the three line segments, we find one true intersection point. Now let us do the same for 5000 line segments. To this end, we generate 20000 numbers using the so-called "Blum Blum Shub" pseudo-random number generator. -s0 = 290797 -sn+1 = snsn (modulo 50515093) +s0 = 290797 +sn+1 = snsn (modulo 50515093) tn = sn (modulo 500) To create each line segment, we use four consecutive numbers tn. That is, the first line segment is given by: (t1, t2) to (t3, t4) diff --git a/project_euler/problems/166_criss_cross.txt b/project_euler/problems/166_criss_cross.txt index 7dcc892..4de361b 100644 --- a/project_euler/problems/166_criss_cross.txt +++ b/project_euler/problems/166_criss_cross.txt @@ -4,10 +4,10 @@ Criss Cross A 4x4 grid is filled with digits d, 0 d 9. It can be seen that in the grid - -6 3 3 0 -5 0 4 3 -0 7 1 4 + +6 3 3 0 +5 0 4 3 +0 7 1 4 1 2 4 5 the sum of each row and each column has the value 12. Moreover the sum of each diagonal is also 12. In how many ways can you fill a 4x4 grid with the digits d, 0 d 9 so that each row, each column, and both diagonals have the same sum? diff --git a/project_euler/problems/167_investigating_ulam_sequences.txt b/project_euler/problems/167_investigating_ulam_sequences.txt index c2ade78..b71913a 100644 --- a/project_euler/problems/167_investigating_ulam_sequences.txt +++ b/project_euler/problems/167_investigating_ulam_sequences.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=167 Investigating Ulam sequences -For two positive integers a and b, the Ulam sequence U(a,b) is defined by U(a,b)1 = a, U(a,b)2 = b and for k > 2, +For two positive integers a and b, the Ulam sequence U(a,b) is defined by U(a,b)1 = a, U(a,b)2 = b and for k > 2, U(a,b)k is the smallest integer greater than U(a,b)(k-1) which can be written in exactly one way as the sum of two distinct previous members of U(a,b). -For example, the sequence U(1,2) begins with -1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8; +For example, the sequence U(1,2) begins with +1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8; 5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4. Find U(2,2n+1)k for 2 n 10, where k = 1011. diff --git a/project_euler/problems/168_number_rotations.txt b/project_euler/problems/168_number_rotations.txt index 05e58e5..68a9924 100644 --- a/project_euler/problems/168_number_rotations.txt +++ b/project_euler/problems/168_number_rotations.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=168 Number Rotations -Consider the number 142857. We can right-rotate this number by moving the last digit (7) to the front of it, giving us 714285. -It can be verified that 714285=5142857. +Consider the number 142857. We can right-rotate this number by moving the last digit (7) to the front of it, giving us 714285. +It can be verified that 714285=5142857. This demonstrates an unusual property of 142857: it is a divisor of its right-rotation. Find the last 5 digits of the sum of all integers n, 10 n 10100, that have this property. diff --git a/project_euler/problems/169_exploring_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt b/project_euler/problems/169_exploring_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt index 4b3b118..163e27f 100644 --- a/project_euler/problems/169_exploring_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt +++ b/project_euler/problems/169_exploring_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt @@ -4,9 +4,9 @@ Exploring the number of different ways a number can be expressed as a sum of pow Define f(0)=1 and f(n) to be the number of different ways n can be expressed as a sum of integer powers of 2 using each power no more than twice. For example, f(10)=5 since there are five different ways to express 10: -1 + 1 + 8 -1 + 1 + 4 + 41 + 1 + 2 + 2 + 4 -2 + 4 + 4 +1 + 1 + 8 +1 + 1 + 4 + 41 + 1 + 2 + 2 + 4 +2 + 4 + 4 2 + 8 What is f(1025)? diff --git a/project_euler/problems/170_find_the_largest_0_to_9_pandigital_that_can_be_formed_by_concatenating_products.txt b/project_euler/problems/170_find_the_largest_0_to_9_pandigital_that_can_be_formed_by_concatenating_products.txt index a19104c..5afa512 100644 --- a/project_euler/problems/170_find_the_largest_0_to_9_pandigital_that_can_be_formed_by_concatenating_products.txt +++ b/project_euler/problems/170_find_the_largest_0_to_9_pandigital_that_can_be_formed_by_concatenating_products.txt @@ -3,7 +3,7 @@ http://projecteuler.net/problem=170 Find the largest 0 to 9 pandigital that can be formed by concatenating products Take the number 6 and multiply it by each of 1273 and 9854: -6 1273 = 7638 +6 1273 = 7638 6 9854 = 59124 By concatenating these products we get the 1 to 9 pandigital 763859124. We will call 763859124 the "concatenated product of 6 and (1273,9854)". Notice too, that the concatenation of the input numbers, 612739854, is also 1 to 9 pandigital. The same can be done for 0 to 9 pandigital numbers. diff --git a/project_euler/problems/171_finding_numbers_for_which_the_sum_of_the_squares_of_the_digits_is_a_square.txt b/project_euler/problems/171_finding_numbers_for_which_the_sum_of_the_squares_of_the_digits_is_a_square.txt index c8dac5f..bae29a3 100644 --- a/project_euler/problems/171_finding_numbers_for_which_the_sum_of_the_squares_of_the_digits_is_a_square.txt +++ b/project_euler/problems/171_finding_numbers_for_which_the_sum_of_the_squares_of_the_digits_is_a_square.txt @@ -3,8 +3,8 @@ http://projecteuler.net/problem=171 Finding numbers for which the sum of the squares of the digits is a square For a positive integer n, let f(n) be the sum of the squares of the digits (in base 10) of n, e.g. -f(3) = 32 = 9, -f(25) = 22 + 52 = 4 + 25 = 29, +f(3) = 32 = 9, +f(25) = 22 + 52 = 4 + 25 = 29, f(442) = 42 + 42 + 22 = 16 + 16 + 4 = 36 Find the last nine digits of the sum of all n, 0 n 1020, such that f(n) is a perfect square. diff --git a/project_euler/problems/175_fractions_involving_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt b/project_euler/problems/175_fractions_involving_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt index 8cc5e58..aad470e 100644 --- a/project_euler/problems/175_fractions_involving_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt +++ b/project_euler/problems/175_fractions_involving_the_number_of_different_ways_a_number_can_be_expressed_as_a_sum_of_powers_of_2.txt @@ -1,15 +1,15 @@ http://projecteuler.net/problem=175 Fractions involving the number of different ways a number can be expressed as a sum of powers of 2 - -Define f(0)=1 and f(n) to be the number of ways to write n as a sum of powers of 2 where no power occurs more than twice. - -For example, f(10)=5 since there are five different ways to express 10:10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1 - -It can be shown that for every fraction p/q (p0, q0) there exists at least one integer n such that f(n)/f(n-1)=p/q. -For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. -The binary expansion of 241 is 11110001. -Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241. -Find the Shortened Binary Expansion of the smallest n for which f(n)/f(n-1)=123456789/987654321. -Give your answer as comma separated integers, without any whitespaces. + +Define f(0)=1 and f(n) to be the number of ways to write n as a sum of powers of 2 where no power occurs more than twice. + +For example, f(10)=5 since there are five different ways to express 10:10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1 + +It can be shown that for every fraction p/q (p0, q0) there exists at least one integer n such that f(n)/f(n-1)=p/q. +For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. +The binary expansion of 241 is 11110001. +Reading this binary number from the most significant bit to the least significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the string 4,3,1 the Shortened Binary Expansion of 241. +Find the Shortened Binary Expansion of the smallest n for which f(n)/f(n-1)=123456789/987654321. +Give your answer as comma separated integers, without any whitespaces. diff --git a/project_euler/problems/178_step_numbers.txt b/project_euler/problems/178_step_numbers.txt index 1e98955..161a33f 100644 --- a/project_euler/problems/178_step_numbers.txt +++ b/project_euler/problems/178_step_numbers.txt @@ -1,12 +1,12 @@ http://projecteuler.net/problem=178 Step Numbers - -Consider the number 45656. -It can be seen that each pair of consecutive digits of 45656 has a difference of one. -A number for which every pair of consecutive digits has a difference of one is called a step number. -A pandigital number contains every decimal digit from 0 to 9 at least once. - -How many pandigital step numbers less than 1040 are there? - + +Consider the number 45656. +It can be seen that each pair of consecutive digits of 45656 has a difference of one. +A number for which every pair of consecutive digits has a difference of one is called a step number. +A pandigital number contains every decimal digit from 0 to 9 at least once. + +How many pandigital step numbers less than 1040 are there? + diff --git a/project_euler/problems/179_consecutive_positive_divisors.txt b/project_euler/problems/179_consecutive_positive_divisors.txt index 60859ba..0a68689 100644 --- a/project_euler/problems/179_consecutive_positive_divisors.txt +++ b/project_euler/problems/179_consecutive_positive_divisors.txt @@ -1,6 +1,6 @@ http://projecteuler.net/problem=179 -Consecutive positive divisors +Consecutive positive divisors Find the number of integers 1 n 107, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15. diff --git a/project_euler/problems/180_rational_zeros_of_a_function_of_three_variables.txt b/project_euler/problems/180_rational_zeros_of_a_function_of_three_variables.txt index b5ec5d9..29d57a2 100644 --- a/project_euler/problems/180_rational_zeros_of_a_function_of_three_variables.txt +++ b/project_euler/problems/180_rational_zeros_of_a_function_of_three_variables.txt @@ -8,9 +8,9 @@ f2,n(x,y,z) = (xy + yz + zx)*(xn-1 + yn-1 zn-1) f3,n(x,y,z) = xyz*(xn-2 + yn-2 zn-2) and their combination fn(x,y,z) = f1,n(x,y,z) + f2,n(x,y,z) f3,n(x,y,z) -We call (x,y,z) a golden triple of order k if x, y, and z are all rational numbers of the form a / b with +We call (x,y,z) a golden triple of order k if x, y, and z are all rational numbers of the form a / b with 0 a b k and there is (at least) one integer n, so that fn(x,y,z) = 0. -Let s(x,y,z) = x + y + z. +Let s(x,y,z) = x + y + z. Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples (x,y,z) of order 35. All the s(x,y,z) and t must be in reduced form. Find u + v. diff --git a/project_euler/problems/182_rsa_encryption.txt b/project_euler/problems/182_rsa_encryption.txt index 735f159..71c95cf 100644 --- a/project_euler/problems/182_rsa_encryption.txt +++ b/project_euler/problems/182_rsa_encryption.txt @@ -3,19 +3,19 @@ http://projecteuler.net/problem=182 RSA encryption The RSA encryption is based on the following procedure: -Generate two distinct primes p and q.Compute n=pq and φ=(p-1)(q-1). +Generate two distinct primes p and q.Compute n=pq and φ=(p-1)(q-1). Find an integer e, 1eφ, such that gcd(e,φ)=1. -A message in this system is a number in the interval [0,n-1]. -A text to be encrypted is then somehow converted to messages (numbers in the interval [0,n-1]). +A message in this system is a number in the interval [0,n-1]. +A text to be encrypted is then somehow converted to messages (numbers in the interval [0,n-1]). To encrypt the text, for each message, m, c=me mod n is calculated. To decrypt the text, the following procedure is needed: calculate d such that ed=1 mod φ, then for each encrypted message, c, calculate m=cd mod n. There exist values of e and m such that me mod n=m.We call messages m for which me mod n=m unconcealed messages. -An issue when choosing e is that there should not be too many unconcealed messages. For instance, let p=19 and q=37. -Then n=19*37=703 and φ=18*36=648. +An issue when choosing e is that there should not be too many unconcealed messages. For instance, let p=19 and q=37. +Then n=19*37=703 and φ=18*36=648. If we choose e=181, then, although gcd(181,648)=1 it turns out that all possible messages -m (0mn-1) are unconcealed when calculating me mod n. -For any valid choice of e there exist some unconcealed messages. +m (0mn-1) are unconcealed when calculating me mod n. +For any valid choice of e there exist some unconcealed messages. It's important that the number of unconcealed messages is at a minimum. -Choose p=1009 and q=3643. +Choose p=1009 and q=3643. Find the sum of all values of e, 1eφ(1009,3643) and gcd(e,φ)=1, so that the number of unconcealed messages for this value of e is at a minimum. diff --git a/project_euler/problems/183_maximum_product_of_parts.txt b/project_euler/problems/183_maximum_product_of_parts.txt index 1228906..37a3d79 100644 --- a/project_euler/problems/183_maximum_product_of_parts.txt +++ b/project_euler/problems/183_maximum_product_of_parts.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=183 Maximum product of parts -Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r. +Let N be a positive integer and let N be split into k equal parts, r = N/k, so that N = r + r + ... + r. Let P be the product of these parts, P = r r ... r = rk. For example, if 11 is split into five equal parts, 11 = 2.2 + 2.2 + 2.2 + 2.2 + 2.2, then P = 2.25 = 51.53632. Let M(N) = Pmax for a given value of N. diff --git a/project_euler/problems/185_number_mind.txt b/project_euler/problems/185_number_mind.txt index 671bbef..bbc5180 100644 --- a/project_euler/problems/185_number_mind.txt +++ b/project_euler/problems/185_number_mind.txt @@ -5,35 +5,35 @@ Number Mind The game Number Mind is a variant of the well known game Master Mind. Instead of coloured pegs, you have to guess a secret sequence of digits. After each guess you're only told in how many places you've guessed the correct digit. So, if the sequence was 1234 and you guessed 2036, you'd be told that you have one correct digit; however, you would NOT be told that you also have another digit in the wrong place. For instance, given the following guesses for a 5-digit secret sequence, -90342 ;2 correct -70794 ;0 correct -39458 ;2 correct -34109 ;1 correct -51545 ;2 correct +90342 ;2 correct +70794 ;0 correct +39458 ;2 correct +34109 ;1 correct +51545 ;2 correct 12531 ;1 correct The correct sequence 39542 is unique. Based on the following guesses, -5616185650518293 ;2 correct -3847439647293047 ;1 correct -5855462940810587 ;3 correct -9742855507068353 ;3 correct -4296849643607543 ;3 correct -3174248439465858 ;1 correct -4513559094146117 ;2 correct -7890971548908067 ;3 correct -8157356344118483 ;1 correct -2615250744386899 ;2 correct -8690095851526254 ;3 correct -6375711915077050 ;1 correct -6913859173121360 ;1 correct -6442889055042768 ;2 correct -2321386104303845 ;0 correct -2326509471271448 ;2 correct -5251583379644322 ;2 correct -1748270476758276 ;3 correct -4895722652190306 ;1 correct -3041631117224635 ;3 correct -1841236454324589 ;3 correct +5616185650518293 ;2 correct +3847439647293047 ;1 correct +5855462940810587 ;3 correct +9742855507068353 ;3 correct +4296849643607543 ;3 correct +3174248439465858 ;1 correct +4513559094146117 ;2 correct +7890971548908067 ;3 correct +8157356344118483 ;1 correct +2615250744386899 ;2 correct +8690095851526254 ;3 correct +6375711915077050 ;1 correct +6913859173121360 ;1 correct +6442889055042768 ;2 correct +2321386104303845 ;0 correct +2326509471271448 ;2 correct +5251583379644322 ;2 correct +1748270476758276 ;3 correct +4895722652190306 ;1 correct +3041631117224635 ;3 correct +1841236454324589 ;3 correct 2659862637316867 ;2 correct Find the unique 16-digit secret sequence. diff --git a/project_euler/problems/186_connectedness_of_a_network.txt b/project_euler/problems/186_connectedness_of_a_network.txt index 94008dd..d8025f8 100644 --- a/project_euler/problems/186_connectedness_of_a_network.txt +++ b/project_euler/problems/186_connectedness_of_a_network.txt @@ -14,7 +14,7 @@ RecNrCallerCalled The telephone number of the caller and the called number in record n are Caller(n) = S2n-1 and Called(n) = S2n where S1,2,3,... come from the "Lagged Fibonacci Generator": -For 1 k 55, Sk = [100003 - 200003k + 300007k3] (modulo 1000000) +For 1 k 55, Sk = [100003 - 200003k + 300007k3] (modulo 1000000) For 56 k, Sk = [Sk-24 + Sk-55] (modulo 1000000) If Caller(n) = Called(n) then the user is assumed to have misdialled and the call fails; otherwise the call is successful. From the start of the records, we say that any pair of users X and Y are friends if X calls Y or vice-versa. Similarly, X is a friend of a friend of Z if X is a friend of Y and Y is a friend of Z; and so on for longer chains. diff --git a/project_euler/problems/187_semiprimes.txt b/project_euler/problems/187_semiprimes.txt index fcb2a71..4f61058 100644 --- a/project_euler/problems/187_semiprimes.txt +++ b/project_euler/problems/187_semiprimes.txt @@ -3,7 +3,7 @@ http://projecteuler.net/problem=187 Semiprimes A composite is a number containing at least two prime factors. For example, 15 = 3 5; 9 = 3 3; 12 = 2 2 3. -There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: +There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26. How many composite integers, n 108, have precisely two, not necessarily distinct, prime factors? diff --git a/project_euler/problems/188_the_hyperexponentiation_of_a_number.txt b/project_euler/problems/188_the_hyperexponentiation_of_a_number.txt index e7bf26b..33206df 100644 --- a/project_euler/problems/188_the_hyperexponentiation_of_a_number.txt +++ b/project_euler/problems/188_the_hyperexponentiation_of_a_number.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=188 The hyperexponentiation of a number -The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ba, is recursively defined by: -a↑↑1 = a, +The hyperexponentiation or tetration of a number a by a positive integer b, denoted by a↑↑b or ba, is recursively defined by: +a↑↑1 = a, a↑↑(k+1) = a(a↑↑k). - + Thus we have e.g. 3↑↑2 = 33 = 27, hence 3↑↑3 = 327 = 7625597484987 and 3↑↑4 is roughly 103.6383346400240996*10^12. Find the last 8 digits of 1777↑↑1855. diff --git a/project_euler/problems/189_tricolouring_a_triangular_grid.txt b/project_euler/problems/189_tricolouring_a_triangular_grid.txt index 8a0283a..01f9a21 100644 --- a/project_euler/problems/189_tricolouring_a_triangular_grid.txt +++ b/project_euler/problems/189_tricolouring_a_triangular_grid.txt @@ -4,7 +4,7 @@ Tri-colouring a triangular grid Consider the following configuration of 64 triangles: -We wish to colour the interior of each triangle with one of three colours: red, green or blue, so that no two neighbouring triangles have the same colour. Such a colouring shall be called valid. Here, two triangles are said to be neighbouring if they share an edge. +We wish to colour the interior of each triangle with one of three colours: red, green or blue, so that no two neighbouring triangles have the same colour. Such a colouring shall be called valid. Here, two triangles are said to be neighbouring if they share an edge. Note: if they only share a vertex, then they are not neighbours. For example, here is a valid colouring of the above grid: diff --git a/project_euler/problems/190_maximising_a_weighted_product.txt b/project_euler/problems/190_maximising_a_weighted_product.txt index 35d6e6d..05f25d8 100644 --- a/project_euler/problems/190_maximising_a_weighted_product.txt +++ b/project_euler/problems/190_maximising_a_weighted_product.txt @@ -1,6 +1,6 @@ http://projecteuler.net/problem=190 -Maximising a weighted product +Maximising a weighted product Let Sm = (x1, x2, ... , xm) be the m-tuple of positive real numbers with x1 + x2 + ... + xm = m for which Pm = x1 * x22 * ... * xmm is maximised. diff --git a/project_euler/problems/191_prize_strings.txt b/project_euler/problems/191_prize_strings.txt index 0cf2af8..b5b6352 100644 --- a/project_euler/problems/191_prize_strings.txt +++ b/project_euler/problems/191_prize_strings.txt @@ -5,10 +5,10 @@ Prize Strings A particular school offers cash rewards to children with good attendance and punctuality. If they are absent for three consecutive days or late on more than one occasion then they forfeit their prize. During an n-day period a trinary string is formed for each child consisting of L's (late), O's (on time), and A's (absent). Although there are eighty-one trinary strings for a 4-day period that can be formed, exactly forty-three strings would lead to a prize: -OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA -OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO -AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL -AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA +OOOO OOOA OOOL OOAO OOAA OOAL OOLO OOLA OAOO OAOA +OAOL OAAO OAAL OALO OALA OLOO OLOA OLAO OLAA AOOO +AOOA AOOL AOAO AOAA AOAL AOLO AOLA AAOO AAOA AAOL +AALO AALA ALOO ALOA ALAO ALAA LOOO LOOA LOAO LOAA LAOO LAOA LAAO How many "prize" strings exist over a 30-day period? diff --git a/project_euler/problems/192_best_approximations.txt b/project_euler/problems/192_best_approximations.txt index 67feb59..266ee8c 100644 --- a/project_euler/problems/192_best_approximations.txt +++ b/project_euler/problems/192_best_approximations.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=192 Best Approximations -Let x be a real number. +Let x be a real number. A best approximation to x for the denominator bound d is a rational number r/s in reduced form, with s d, such that any rational number which is closer to x than r/s has a denominator larger than d: |p/q-x| |r/s-x| q d For example, the best approximation to 13 for the denominator bound 20 is 18/5 and the best approximation to 13 for the denominator bound 30 is 101/28. diff --git a/project_euler/problems/194_coloured_configurations.txt b/project_euler/problems/194_coloured_configurations.txt index 67e789f..d74e4cd 100644 --- a/project_euler/problems/194_coloured_configurations.txt +++ b/project_euler/problems/194_coloured_configurations.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=194 Coloured Configurations -Consider graphs built with the units A: -and B: , where the units are glued along +Consider graphs built with the units A: +and B: , where the units are glued along the vertical edges as in the graph . -A configuration of type (a,b,c) is a graph thus built of a units A and b units B, where the graph's vertices are coloured using up to c colours, so that no two adjacent vertices have the same colour. +A configuration of type (a,b,c) is a graph thus built of a units A and b units B, where the graph's vertices are coloured using up to c colours, so that no two adjacent vertices have the same colour. The compound graph above is an example of a configuration of type (2,2,6), in fact of type (2,2,c) for all c 4. -Let N(a,b,c) be the number of configurations of type (a,b,c). +Let N(a,b,c) be the number of configurations of type (a,b,c). For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736. Find the last 8 digits of N(25,75,1984). diff --git a/project_euler/problems/195_inscribed_circles_of_triangles_with_one_angle_of_60_degrees.txt b/project_euler/problems/195_inscribed_circles_of_triangles_with_one_angle_of_60_degrees.txt index 5350356..02f4c54 100644 --- a/project_euler/problems/195_inscribed_circles_of_triangles_with_one_angle_of_60_degrees.txt +++ b/project_euler/problems/195_inscribed_circles_of_triangles_with_one_angle_of_60_degrees.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=195 Inscribed circles of triangles with one angle of 60 degrees -Let's call an integer sided triangle with exactly one angle of 60 degrees a 60-degree triangle. +Let's call an integer sided triangle with exactly one angle of 60 degrees a 60-degree triangle. Let r be the radius of the inscribed circle of such a 60-degree triangle. -There are 1234 60-degree triangles for which r 100. -Let T(n) be the number of 60-degree triangles for which r n, so +There are 1234 60-degree triangles for which r 100. +Let T(n) be the number of 60-degree triangles for which r n, so T(100) = 1234,  T(1000) = 22767, and  T(10000) = 359912. Find T(1053779). diff --git a/project_euler/problems/196_prime_triplets.txt b/project_euler/problems/196_prime_triplets.txt index f324002..5291f34 100644 --- a/project_euler/problems/196_prime_triplets.txt +++ b/project_euler/problems/196_prime_triplets.txt @@ -3,24 +3,24 @@ http://projecteuler.net/problem=196 Prime triplets Build a triangle from all positive integers in the following way: - 1 - 2  3 - 4  5  6 + 1 + 2  3 + 4  5  6  7  8  9 10 -11 12 13 14 15 -16 17 18 19 20 21 +11 12 13 14 15 +16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 -37 38 39 40 41 42 43 44 45 -46 47 48 49 50 51 52 53 54 55 -56 57 58 59 60 61 62 63 64 65 66 +37 38 39 40 41 42 43 44 45 +46 47 48 49 50 51 52 53 54 55 +56 57 58 59 60 61 62 63 64 65 66 . . . Each positive integer has up to eight neighbours in the triangle. A set of three primes is called a prime triplet if one of the three primes has the other two as neighbours in the triangle. For example, in the second row, the prime numbers 2 and 3 are elements of some prime triplet. -If row 8 is considered, it contains two primes which are elements of some prime triplet, i.e. 29 and 31. +If row 8 is considered, it contains two primes which are elements of some prime triplet, i.e. 29 and 31. If row 9 is considered, it contains only one prime which is an element of some prime triplet: 37. -Define S(n) as the sum of the primes in row n which are elements of any prime triplet. +Define S(n) as the sum of the primes in row n which are elements of any prime triplet. Then S(8)=60 and S(9)=37. You are given that S(10000)=950007619. Find  S(5678027) + S(7208785). diff --git a/project_euler/problems/197_investigating_the_behaviour_of_a_recursively_defined_sequence.txt b/project_euler/problems/197_investigating_the_behaviour_of_a_recursively_defined_sequence.txt index 3eaf1c5..9e7424e 100644 --- a/project_euler/problems/197_investigating_the_behaviour_of_a_recursively_defined_sequence.txt +++ b/project_euler/problems/197_investigating_the_behaviour_of_a_recursively_defined_sequence.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=197 Investigating the behaviour of a recursively defined sequence -Given is the function f(x) = 230.403243784-x2 10-9 ( is the floor-function), +Given is the function f(x) = 230.403243784-x2 10-9 ( is the floor-function), the sequence un is defined by u0 = -1 and un+1 = f(un). -Find un + un+1 for n = 1012. +Find un + un+1 for n = 1012. Give your answer with 9 digits after the decimal point. diff --git a/project_euler/problems/198_ambiguous_numbers.txt b/project_euler/problems/198_ambiguous_numbers.txt index 9de678b..ddb1d2d 100644 --- a/project_euler/problems/198_ambiguous_numbers.txt +++ b/project_euler/problems/198_ambiguous_numbers.txt @@ -3,8 +3,8 @@ http://projecteuler.net/problem=198 Ambiguous Numbers A best approximation to a real number x for the denominator bound d is a rational number r/s (in reduced form) with s d, so that any rational number p/q which is closer to x than r/s has q d. -Usually the best approximation to a real number is uniquely determined for all denominator bounds. However, there are some exceptions, e.g. 9/40 has the two best approximations 1/4 and 1/5 for the denominator bound 6. +Usually the best approximation to a real number is uniquely determined for all denominator bounds. However, there are some exceptions, e.g. 9/40 has the two best approximations 1/4 and 1/5 for the denominator bound 6. We shall call a real number x ambiguous, if there is at least one denominator bound for which x possesses two best approximations. Clearly, an ambiguous number is necessarily rational. -How many ambiguous numbers x = p/q, +How many ambiguous numbers x = p/q, 0 x 1/100, are there whose denominator q does not exceed 108? diff --git a/project_euler/problems/199_iterative_circle_packing.txt b/project_euler/problems/199_iterative_circle_packing.txt index a0e329b..5790e90 100644 --- a/project_euler/problems/199_iterative_circle_packing.txt +++ b/project_euler/problems/199_iterative_circle_packing.txt @@ -6,11 +6,11 @@ Three circles of equal radius are placed inside a larger circle such that each p - -At each iteration, a maximally sized circle is placed in each gap, which creates more gaps for the next iteration. After 3 iterations (pictured), there are 108 gaps and the fraction of the area which is not covered by circles is 0.06790342, rounded to eight decimal places. - -What fraction of the area is not covered by circles after 10 iterations? -Give your answer rounded to eight decimal places using the format x.xxxxxxxx . +At each iteration, a maximally sized circle is placed in each gap, which creates more gaps for the next iteration. After 3 iterations (pictured), there are 108 gaps and the fraction of the area which is not covered by circles is 0.06790342, rounded to eight decimal places. + + +What fraction of the area is not covered by circles after 10 iterations? +Give your answer rounded to eight decimal places using the format x.xxxxxxxx . diff --git a/project_euler/problems/200_find_the_200th_primeproof_sqube_containing_the_contiguous_substring_200.txt b/project_euler/problems/200_find_the_200th_primeproof_sqube_containing_the_contiguous_substring_200.txt index ff8b863..54f6e8b 100644 --- a/project_euler/problems/200_find_the_200th_primeproof_sqube_containing_the_contiguous_substring_200.txt +++ b/project_euler/problems/200_find_the_200th_primeproof_sqube_containing_the_contiguous_substring_200.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=200 Find the 200th prime-proof sqube containing the contiguous sub-string "200" -We shall define a sqube to be a number of the form, p2q3, where p and q are distinct primes. +We shall define a sqube to be a number of the form, p2q3, where p and q are distinct primes. For example, 200 = 5223 or 120072949 = 232613. The first five squbes are 72, 108, 200, 392, and 500. Interestingly, 200 is also the first number for which you cannot change any single digit to make a prime; we shall call such numbers, prime-proof. The next prime-proof sqube which contains the contiguous sub-string "200" is 1992008. diff --git a/project_euler/problems/201_subsets_with_a_unique_sum.txt b/project_euler/problems/201_subsets_with_a_unique_sum.txt index abfe3d4..f07ccf5 100644 --- a/project_euler/problems/201_subsets_with_a_unique_sum.txt +++ b/project_euler/problems/201_subsets_with_a_unique_sum.txt @@ -2,32 +2,32 @@ http://projecteuler.net/problem=201 Subsets with a unique sum -For any set A of numbers, let sum(A) be the sum of the elements of A. +For any set A of numbers, let sum(A) be the sum of the elements of A. Consider the set B = {1,3,6,8,10,11}. There are 20 subsets of B containing three elements, and their sums are: - -sum({1,3,6}) = 10, -sum({1,3,8}) = 12, -sum({1,3,10}) = 14, -sum({1,3,11}) = 15, -sum({1,6,8}) = 15, -sum({1,6,10}) = 17, -sum({1,6,11}) = 18, -sum({1,8,10}) = 19, -sum({1,8,11}) = 20, -sum({1,10,11}) = 22, -sum({3,6,8}) = 17, -sum({3,6,10}) = 19, -sum({3,6,11}) = 20, -sum({3,8,10}) = 21, -sum({3,8,11}) = 22, -sum({3,10,11}) = 24, -sum({6,8,10}) = 24, -sum({6,8,11}) = 25, -sum({6,10,11}) = 27, + +sum({1,3,6}) = 10, +sum({1,3,8}) = 12, +sum({1,3,10}) = 14, +sum({1,3,11}) = 15, +sum({1,6,8}) = 15, +sum({1,6,10}) = 17, +sum({1,6,11}) = 18, +sum({1,8,10}) = 19, +sum({1,8,11}) = 20, +sum({1,10,11}) = 22, +sum({3,6,8}) = 17, +sum({3,6,10}) = 19, +sum({3,6,11}) = 20, +sum({3,8,10}) = 21, +sum({3,8,11}) = 22, +sum({3,10,11}) = 24, +sum({6,8,10}) = 24, +sum({6,8,11}) = 25, +sum({6,10,11}) = 27, sum({8,10,11}) = 29. -Some of these sums occur more than once, others are unique. +Some of these sums occur more than once, others are unique. For a set A, let U(A,k) be the set of unique sums of k-element subsets of A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and sum(U(B,3)) = 156. -Now consider the 100-element set S = {12, 22, ... , 1002}. +Now consider the 100-element set S = {12, 22, ... , 1002}. S has 100891344545564193334812497256 50-element subsets. Determine the sum of all integers which are the sum of exactly one of the 50-element subsets of S, i.e. find sum(U(S,50)). diff --git a/project_euler/problems/203_squarefree_binomial_coefficients.txt b/project_euler/problems/203_squarefree_binomial_coefficients.txt index 2828e98..8312c9d 100644 --- a/project_euler/problems/203_squarefree_binomial_coefficients.txt +++ b/project_euler/problems/203_squarefree_binomial_coefficients.txt @@ -13,12 +13,12 @@ The binomial coefficients nCk can be arranged in triangular form, Pascal's trian 15101051 1615201561 172135352171 - -......... + +......... It can be seen that the first eight rows of Pascal's triangle contain twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35. -A positive integer n is called squarefree if no square of a prime divides n. -Of the twelve distinct numbers in the first eight rows of Pascal's triangle, all except 4 and 20 are squarefree. +A positive integer n is called squarefree if no square of a prime divides n. +Of the twelve distinct numbers in the first eight rows of Pascal's triangle, all except 4 and 20 are squarefree. The sum of the distinct squarefree numbers in the first eight rows is 105. Find the sum of the distinct squarefree numbers in the first 51 rows of Pascal's triangle. diff --git a/project_euler/problems/204_generalised_hamming_numbers.txt b/project_euler/problems/204_generalised_hamming_numbers.txt index 529b3e3..addc6c6 100644 --- a/project_euler/problems/204_generalised_hamming_numbers.txt +++ b/project_euler/problems/204_generalised_hamming_numbers.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=204 Generalised Hamming Numbers -A Hamming number is a positive number which has no prime factor larger than 5. -So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15. +A Hamming number is a positive number which has no prime factor larger than 5. +So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15. There are 1105 Hamming numbers not exceeding 108. -We will call a positive number a generalised Hamming number of type n, if it has no prime factor larger than n. +We will call a positive number a generalised Hamming number of type n, if it has no prime factor larger than n. Hence the Hamming numbers are the generalised Hamming numbers of type 5. How many generalised Hamming numbers of type 100 are there which don't exceed 109? diff --git a/project_euler/problems/205_dice_game.txt b/project_euler/problems/205_dice_game.txt index 2cc36eb..237b7ad 100644 --- a/project_euler/problems/205_dice_game.txt +++ b/project_euler/problems/205_dice_game.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=205 Dice Game -Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4. +Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4. Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5, 6. Peter and Colin roll their dice and compare totals: the highest total wins. The result is a draw if the totals are equal. What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer rounded to seven decimal places in the form 0.abcdefg diff --git a/project_euler/problems/207_integer_partition_equations.txt b/project_euler/problems/207_integer_partition_equations.txt index e456437..0a65163 100644 --- a/project_euler/problems/207_integer_partition_equations.txt +++ b/project_euler/problems/207_integer_partition_equations.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=207 Integer partition equations -For some positive integers k, there exists an integer partition of the form   4t = 2t + k, +For some positive integers k, there exists an integer partition of the form   4t = 2t + k, where 4t, 2t, and k are all positive integers and t is a real number. The first two such partitions are 41 = 21 + 2 and 41.5849625... = 21.5849625... + 6. -Partitions where t is also an integer are called perfect. -For any m 1 let P(m) be the proportion of such partitions that are perfect with k m. +Partitions where t is also an integer are called perfect. +For any m 1 let P(m) be the proportion of such partitions that are perfect with k m. Thus P(6) = 1/2. In the following table are listed some values of P(m) -   P(5) = 1/1 -   P(10) = 1/2 -   P(15) = 2/3 -   P(20) = 1/2 -   P(25) = 1/2 -   P(30) = 2/5 -   ... -   P(180) = 1/4 +   P(5) = 1/1 +   P(10) = 1/2 +   P(15) = 2/3 +   P(20) = 1/2 +   P(25) = 1/2 +   P(30) = 2/5 +   ... +   P(180) = 1/4    P(185) = 3/13 Find the smallest m for which P(m) 1/12345 diff --git a/project_euler/problems/208_robot_walks.txt b/project_euler/problems/208_robot_walks.txt index 84b74a2..ed4c68d 100644 --- a/project_euler/problems/208_robot_walks.txt +++ b/project_euler/problems/208_robot_walks.txt @@ -8,7 +8,7 @@ One of 70932 possible closed paths of 25 arcs starting northward is Given that the robot starts facing North, how many journeys of 70 arcs in length can it take that return it, after the final arc, to its starting position? - -(Any arc may be traversed multiple times.) + +(Any arc may be traversed multiple times.) diff --git a/project_euler/problems/209_circular_logic.txt b/project_euler/problems/209_circular_logic.txt index f772c04..9706ffb 100644 --- a/project_euler/problems/209_circular_logic.txt +++ b/project_euler/problems/209_circular_logic.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=209 Circular Logic -A k-input binary truth table is a map from k input bits +A k-input binary truth table is a map from k input bits (binary digits, 0 [false] or 1 [true]) to 1 output bit. For example, the 2-input binary truth tables for the logical AND and XOR functions are: @@ -27,9 +27,9 @@ x XOR y How many 6-input binary truth tables, τ, satisfy the formula - -τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0 -for all 6-bit inputs (a, b, c, d, e, f)? +τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0 + +for all 6-bit inputs (a, b, c, d, e, f)? diff --git a/project_euler/problems/210_obtuse_angled_triangles.txt b/project_euler/problems/210_obtuse_angled_triangles.txt index 9a3920a..1459175 100644 --- a/project_euler/problems/210_obtuse_angled_triangles.txt +++ b/project_euler/problems/210_obtuse_angled_triangles.txt @@ -1,12 +1,12 @@ http://projecteuler.net/problem=210 Obtuse Angled Triangles - -Consider the set S(r) of points (x,y) with integer coordinates satisfying |x| + |y| r. -Let O be the point (0,0) and C the point (r/4,r/4). -Let N(r) be the number of points B in S(r), so that the triangle OBC has an obtuse angle, i.e. the largest angle α satisfies 90° -So, for example, N(4)=24 and N(8)=100. - -What is N(1,000,000,000)? + +Consider the set S(r) of points (x,y) with integer coordinates satisfying |x| + |y| r. +Let O be the point (0,0) and C the point (r/4,r/4). +Let N(r) be the number of points B in S(r), so that the triangle OBC has an obtuse angle, i.e. the largest angle α satisfies 90° +So, for example, N(4)=24 and N(8)=100. + +What is N(1,000,000,000)? diff --git a/project_euler/problems/213_flea_circus.txt b/project_euler/problems/213_flea_circus.txt index 5d88021..1d097a9 100644 --- a/project_euler/problems/213_flea_circus.txt +++ b/project_euler/problems/213_flea_circus.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=213 Flea Circus -A 3030 grid of squares contains 900 fleas, initially one flea per square. +A 3030 grid of squares contains 900 fleas, initially one flea per square. When a bell is rung, each flea jumps to an adjacent square at random (usually 4 possibilities, except for fleas on the edge of the grid or at the corners). What is the expected number of unoccupied squares after 50 rings of the bell? Give your answer rounded to six decimal places. diff --git a/project_euler/problems/214_totient_chains.txt b/project_euler/problems/214_totient_chains.txt index 5dcb7e9..954f566 100644 --- a/project_euler/problems/214_totient_chains.txt +++ b/project_euler/problems/214_totient_chains.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=214 Totient Chains -Let φ be Euler's totient function, i.e. for a natural number n, +Let φ be Euler's totient function, i.e. for a natural number n, φ(n) is the number of k, 1 k n, for which gcd(k,n) = 1. -By iterating φ, each positive integer generates a decreasing chain of numbers ending in 1. -E.g. if we start with 5 the sequence 5,4,2,1 is generated. +By iterating φ, each positive integer generates a decreasing chain of numbers ending in 1. +E.g. if we start with 5 the sequence 5,4,2,1 is generated. Here is a listing of all chains with length 4: - -5,4,2,1 -7,6,2,1 -8,4,2,1 -9,6,2,1 -10,4,2,1 -12,4,2,1 -14,6,2,1 + +5,4,2,1 +7,6,2,1 +8,4,2,1 +9,6,2,1 +10,4,2,1 +12,4,2,1 +14,6,2,1 18,6,2,1 Only two of these chains start with a prime, their sum is 12. What is the sum of all primes less than 40000000 which generate a chain of length 25? diff --git a/project_euler/problems/216_investigating_the_primality_of_numbers_of_the_form_2n21.txt b/project_euler/problems/216_investigating_the_primality_of_numbers_of_the_form_2n21.txt index c1fd8d2..89c69b8 100644 --- a/project_euler/problems/216_investigating_the_primality_of_numbers_of_the_form_2n21.txt +++ b/project_euler/problems/216_investigating_the_primality_of_numbers_of_the_form_2n21.txt @@ -2,9 +2,9 @@ http://projecteuler.net/problem=216 Investigating the primality of numbers of the form 2n2-1 -Consider numbers t(n) of the form t(n) = 2n2-1 with n 1. -The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161. -It turns out that only 49 = 7*7 and 161 = 7*23 are not prime. +Consider numbers t(n) of the form t(n) = 2n2-1 with n 1. +The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161. +It turns out that only 49 = 7*7 and 161 = 7*23 are not prime. For n 10000 there are 2202 numbers t(n) that are prime. How many numbers t(n) are prime for n 50,000,000 ? diff --git a/project_euler/problems/217_balanced_numbers.txt b/project_euler/problems/217_balanced_numbers.txt index b16dbb4..596f60e 100644 --- a/project_euler/problems/217_balanced_numbers.txt +++ b/project_euler/problems/217_balanced_numbers.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=217 Balanced Numbers - + A positive integer with k (decimal) digits is called balanced if its first k/2 digits sum to the same value as its last k/2 digits, where x, pronounced ceiling of x, is the smallest integer x, thus π = 4 and 5 = 5. So, for example, all palindromes are balanced, as is 13722. -Let T(n) be the sum of all balanced numbers less than 10n. -Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890. +Let T(n) be the sum of all balanced numbers less than 10n. +Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890. Find T(47) mod 315 diff --git a/project_euler/problems/218_perfect_rightangled_triangles.txt b/project_euler/problems/218_perfect_rightangled_triangles.txt index c4a4317..e19bbe0 100644 --- a/project_euler/problems/218_perfect_rightangled_triangles.txt +++ b/project_euler/problems/218_perfect_rightangled_triangles.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=218 Perfect right-angled triangles -Consider the right angled triangle with sides a=7, b=24 and c=25. -The area of this triangle is 84, which is divisible by the perfect numbers 6 and 28. -Moreover it is a primitive right angled triangle as gcd(a,b)=1 and gcd(b,c)=1. +Consider the right angled triangle with sides a=7, b=24 and c=25. +The area of this triangle is 84, which is divisible by the perfect numbers 6 and 28. +Moreover it is a primitive right angled triangle as gcd(a,b)=1 and gcd(b,c)=1. Also c is a perfect square. -We will call a right angled triangle perfect if --it is a primitive right angled triangle +We will call a right angled triangle perfect if +-it is a primitive right angled triangle -its hypotenuse is a perfect square -We will call a right angled triangle super-perfect if --it is a perfect right angled triangle and --its area is a multiple of the perfect numbers 6 and 28. +We will call a right angled triangle super-perfect if +-it is a perfect right angled triangle and +-its area is a multiple of the perfect numbers 6 and 28. How many perfect right-angled triangles with c1016 exist that are not super-perfect? diff --git a/project_euler/problems/219_skewcost_coding.txt b/project_euler/problems/219_skewcost_coding.txt index 47ccc7a..a093ca5 100644 --- a/project_euler/problems/219_skewcost_coding.txt +++ b/project_euler/problems/219_skewcost_coding.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=219 Skew-cost coding -Let A and B be bit strings (sequences of 0's and 1's). -If A is equal to the leftmost length(A) bits of B, then A is said to be a prefix of B. +Let A and B be bit strings (sequences of 0's and 1's). +If A is equal to the leftmost length(A) bits of B, then A is said to be a prefix of B. For example, 00110 is a prefix of 001101001, but not of 00111 or 100110. A prefix-free code of size n is a collection of n distinct bit strings such that no string is a prefix of any other. For example, this is a prefix-free code of size 6: 0000, 0001, 001, 01, 10, 11 -Now suppose that it costs one penny to transmit a '0' bit, but four pence to transmit a '1'. -Then the total cost of the prefix-free code shown above is 35 pence, which happens to be the cheapest possible for the skewed pricing scheme in question. +Now suppose that it costs one penny to transmit a '0' bit, but four pence to transmit a '1'. +Then the total cost of the prefix-free code shown above is 35 pence, which happens to be the cheapest possible for the skewed pricing scheme in question. In short, we write Cost(6) = 35. What is Cost(109) ? diff --git a/project_euler/problems/220_heighway_dragon.txt b/project_euler/problems/220_heighway_dragon.txt index e67da24..393e6a5 100644 --- a/project_euler/problems/220_heighway_dragon.txt +++ b/project_euler/problems/220_heighway_dragon.txt @@ -3,13 +3,13 @@ http://projecteuler.net/problem=220 Heighway Dragon Let D0 be the two-letter string "Fa". For n1, derive Dn from Dn-1 by the string-rewriting rules: -"a" "aRbFR" +"a" "aRbFR" "b" "LFaLb" Thus, D0 = "Fa", D1 = "FaRbFR", D2 = "FaRbFRRLFaLbFR", and so on. These strings can be interpreted as instructions to a computer graphics program, with "F" meaning "draw forward one unit", "L" meaning "turn left 90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being ignored. The initial position of the computer cursor is (0,0), pointing up towards (0,1). Then Dn is an exotic drawing known as the Heighway Dragon of order n. For example, D10 is shown below; counting each "F" as one step, the highlighted spot at (18,16) is the position reached after 500 steps. -What is the position of the cursor after 1012 steps in D50 ? +What is the position of the cursor after 1012 steps in D50 ? Give your answer in the form x,y with no spaces. diff --git a/project_euler/problems/221_alexandrian_integers.txt b/project_euler/problems/221_alexandrian_integers.txt index df63514..a67a1b5 100644 --- a/project_euler/problems/221_alexandrian_integers.txt +++ b/project_euler/problems/221_alexandrian_integers.txt @@ -6,7 +6,7 @@ We shall call a positive integer A an "Alexandrian integer", if there exist inte -A = p · q · r    and   +A = p · q · r    and   @@ -37,7 +37,7 @@ r -For example, 630 is an Alexandrian integer (p = 5, q = 7, r = 18). +For example, 630 is an Alexandrian integer (p = 5, q = 7, r = 18). In fact, 630 is the 6th Alexandrian integer, the first 6 Alexandrian integers being: 6, 42, 120, 156, 420 and 630. Find the 150000th Alexandrian integer. diff --git a/project_euler/problems/225_tribonacci_nondivisors.txt b/project_euler/problems/225_tribonacci_nondivisors.txt index fc337fc..380ef5e 100644 --- a/project_euler/problems/225_tribonacci_nondivisors.txt +++ b/project_euler/problems/225_tribonacci_nondivisors.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=225 Tribonacci non-divisors - -The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ... -is defined by T1 = T2 = T3 = 1 and Tn = Tn-1 + Tn-2 + Tn-3. - +The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ... +is defined by T1 = T2 = T3 = 1 and Tn = Tn-1 + Tn-2 + Tn-3. + + It can be shown that 27 does not divide any terms of this sequence.In fact, 27 is the first odd number with this property. - + Find the 124th odd number that does not divide any terms of the above sequence. diff --git a/project_euler/problems/227_the_chase.txt b/project_euler/problems/227_the_chase.txt index 261ff7b..72e7238 100644 --- a/project_euler/problems/227_the_chase.txt +++ b/project_euler/problems/227_the_chase.txt @@ -3,8 +3,8 @@ http://projecteuler.net/problem=227 The Chase "The Chase" is a game played with two dice and an even number of players. -The players sit around a table; the game begins with two opposite players having one die each. On each turn, the two players with a die roll it. -If a player rolls a 1, he passes the die to his neighbour on the left; if he rolls a 6, he passes the die to his neighbour on the right; otherwise, he keeps the die for the next turn. +The players sit around a table; the game begins with two opposite players having one die each. On each turn, the two players with a die roll it. +If a player rolls a 1, he passes the die to his neighbour on the left; if he rolls a 6, he passes the die to his neighbour on the right; otherwise, he keeps the die for the next turn. The game ends when one player has both dice after they have been rolled and passed; that player has then lost. In a game with 100 players, what is the expected number of turns the game lasts? Give your answer rounded to ten significant digits. diff --git a/project_euler/problems/228_minkowski_sums.txt b/project_euler/problems/228_minkowski_sums.txt index 01cb279..5c5cd71 100644 --- a/project_euler/problems/228_minkowski_sums.txt +++ b/project_euler/problems/228_minkowski_sums.txt @@ -2,25 +2,25 @@ http://projecteuler.net/problem=228 Minkowski Sums -Let Sn be the regular n-sided polygon – or shape – whose vertices - +Let Sn be the regular n-sided polygon – or shape – whose vertices + vk (k = 1,2,…,n) have coordinates: -xk   =   +xk   =   cos( 2k-1/n 180° ) -yk   =   +yk   =   sin( 2k-1/n 180° ) Each Sn is to be interpreted as a filled shape consisting of all points on the perimeter and in the interior. -The Minkowski sum, S+T, of two shapes S and T is the result of - -adding every point in S to every point in T, where point addition is performed coordinate-wise: - +The Minkowski sum, S+T, of two shapes S and T is the result of + +adding every point in S to every point in T, where point addition is performed coordinate-wise: + (u, v) + (x, y) = (u+x, v+y). For example, the sum of S3 and S4 is the six-sided shape shown in pink below: diff --git a/project_euler/problems/229_four_representations_using_squares.txt b/project_euler/problems/229_four_representations_using_squares.txt index c247ec7..a72b956 100644 --- a/project_euler/problems/229_four_representations_using_squares.txt +++ b/project_euler/problems/229_four_representations_using_squares.txt @@ -3,23 +3,23 @@ http://projecteuler.net/problem=229 Four Representations using Squares Consider the number 3600. It is very special, because - -3600 = 482 +     362 -3600 = 202 + 2402 -3600 = 302 + 3302 + +3600 = 482 +     362 +3600 = 202 + 2402 +3600 = 302 + 3302 3600 = 452 + 7152 Similarly, we find that 88201 = 992 + 2802 = 2872 + 2542 = 2832 + 3522 = 1972 + 7842. -In 1747, Euler proved which numbers are representable as a sum of two squares. +In 1747, Euler proved which numbers are representable as a sum of two squares. We are interested in the numbers n which admit representations of all of the following four types: n = a12 +   b12 n = a22 + 2 b22 n = a32 + 3 b32 -n = a72 + 7 b72, +n = a72 + 7 b72, where the ak and bk are positive integers. -There are 75373 such numbers that do not exceed 107. - +There are 75373 such numbers that do not exceed 107. + How many such numbers are there that do not exceed 2109? diff --git a/project_euler/problems/230_fibonacci_words.txt b/project_euler/problems/230_fibonacci_words.txt index 8b777ee..e24d60c 100644 --- a/project_euler/problems/230_fibonacci_words.txt +++ b/project_euler/problems/230_fibonacci_words.txt @@ -6,19 +6,19 @@ For any two strings of digits, A and B, we define FA,B to be the sequence (A,B,A Further, we define DA,B(n) to be the nth digit in the first term of FA,B that contains at least n digits. Example: Let A=1415926535, B=8979323846. We wish to find DA,B(35), say. -The first few terms of FA,B are: -1415926535 -8979323846 -14159265358979323846 -897932384614159265358979323846 +The first few terms of FA,B are: +1415926535 +8979323846 +14159265358979323846 +897932384614159265358979323846 14159265358979323846897932384614159265358979323846 Then DA,B(35) is the 35th digit in the fifth term, which is 9. Now we use for A the first 100 digits of π behind the decimal point: -14159265358979323846264338327950288419716939937510 -58209749445923078164062862089986280348253421170679 +14159265358979323846264338327950288419716939937510 +58209749445923078164062862089986280348253421170679 and for B the next hundred digits: -82148086513282306647093844609550582231725359408128 +82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 . Find n = 0,1,...,17   10n DA,B((127+19n)7n) . diff --git a/project_euler/problems/231_the_prime_factorisation_of_binomial_coefficients.txt b/project_euler/problems/231_the_prime_factorisation_of_binomial_coefficients.txt index a0c94cb..f5bf101 100644 --- a/project_euler/problems/231_the_prime_factorisation_of_binomial_coefficients.txt +++ b/project_euler/problems/231_the_prime_factorisation_of_binomial_coefficients.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=231 The prime factorisation of binomial coefficients -The binomial coefficient 10C3 = 120. -120 = 23 3 5 = 2 2 2 3 5, and 2 + 2 + 2 + 3 + 5 = 14. -So the sum of the terms in the prime factorisation of 10C3 is 14. - -Find the sum of the terms in the prime factorisation of 20000000C15000000. +The binomial coefficient 10C3 = 120. +120 = 23 3 5 = 2 2 2 3 5, and 2 + 2 + 2 + 3 + 5 = 14. +So the sum of the terms in the prime factorisation of 10C3 is 14. + +Find the sum of the terms in the prime factorisation of 20000000C15000000. diff --git a/project_euler/problems/234_semidivisible_numbers.txt b/project_euler/problems/234_semidivisible_numbers.txt index 229291b..0ae10b5 100644 --- a/project_euler/problems/234_semidivisible_numbers.txt +++ b/project_euler/problems/234_semidivisible_numbers.txt @@ -3,9 +3,9 @@ http://projecteuler.net/problem=234 Semidivisible numbers For an integer n 4, we define the lower prime square root of n, denoted by lps(n), as the largest prime n and the upper prime square root of n, ups(n), as the smallest prime n. -So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. +So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. Let us call an integer n 4 semidivisible, if one of lps(n) and ups(n) divides n, but not both. -The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8, 10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3 and ups(15) = 5. +The sum of the semidivisible numbers not exceeding 15 is 30, the numbers are 8, 10 and 12. 15 is not semidivisible because it is a multiple of both lps(15) = 3 and ups(15) = 5. As a further example, the sum of the 92 semidivisible numbers up to 1000 is 34825. What is the sum of all semidivisible numbers not exceeding 999966663333 ? diff --git a/project_euler/problems/235_an_arithmetic_geometric_sequence.txt b/project_euler/problems/235_an_arithmetic_geometric_sequence.txt index 11e4d04..c4cf103 100644 --- a/project_euler/problems/235_an_arithmetic_geometric_sequence.txt +++ b/project_euler/problems/235_an_arithmetic_geometric_sequence.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=235 An Arithmetic Geometric sequence - -Given is the arithmetic-geometric sequence u(k) = (900-3k)rk-1. -Let s(n) = Σk=1...nu(k). - -Find the value of r for which s(5000) = -600,000,000,000. +Given is the arithmetic-geometric sequence u(k) = (900-3k)rk-1. +Let s(n) = Σk=1...nu(k). - -Give your answer rounded to 12 places behind the decimal point. + +Find the value of r for which s(5000) = -600,000,000,000. + + +Give your answer rounded to 12 places behind the decimal point. diff --git a/project_euler/problems/236_luxury_hampers.txt b/project_euler/problems/236_luxury_hampers.txt index 6777734..455f133 100644 --- a/project_euler/problems/236_luxury_hampers.txt +++ b/project_euler/problems/236_luxury_hampers.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=236 Luxury Hampers - -table.p236, table.p236 th, table.p236 td { - border-width: 1px 1px 1px 1px; - border-style: solid solid solid solid; - border-color: black black black black; - text-align:right; - -moz-border-radius: 0px 0px 0px 0px; -} -table.p236 { - border-spacing: 1px; - border-collapse: separate; - background-color: rgb(224,237,252); -} -table.p236 th, table.p236 td { - padding: 1px 6px 1px 6px; -} -table.p236 th { background-color: rgb(193,218,249); } -table.p236 td { background-color: rgb(255,255,255); } + +table.p236, table.p236 th, table.p236 td { + border-width: 1px 1px 1px 1px; + border-style: solid solid solid solid; + border-color: black black black black; + text-align:right; + -moz-border-radius: 0px 0px 0px 0px; +} +table.p236 { + border-spacing: 1px; + border-collapse: separate; + background-color: rgb(224,237,252); +} +table.p236 th, table.p236 td { + padding: 1px 6px 1px 6px; +} +table.p236 th { background-color: rgb(193,218,249); } +table.p236 td { background-color: rgb(255,255,255); } Suppliers 'A' and 'B' provided the following numbers of products for the luxury hamper market: @@ -36,6 +36,6 @@ The five per-product spoilage rates for each supplier are equal to the number of The overall spoilage rate for each supplier is equal to the total number of products gone bad divided by the total number of products provided by that supplier. To their surprise, the suppliers found that each of the five per-product spoilage rates was worse (higher) for 'B' than for 'A' by the same factor (ratio of spoilage rates), m>1; and yet, paradoxically, the overall spoilage rate was worse for 'A' than for 'B', also by a factor of m. There are thirty-five m1 for which this surprising result could have occurred, the smallest of which is 1476/1475. -What's the largest possible value of m? +What's the largest possible value of m? Give your answer as a fraction reduced to its lowest terms, in the form u/v. diff --git a/project_euler/problems/238_infinite_string_tour.txt b/project_euler/problems/238_infinite_string_tour.txt index d157eae..de73fbc 100644 --- a/project_euler/problems/238_infinite_string_tour.txt +++ b/project_euler/problems/238_infinite_string_tour.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=238 Infinite string tour - -table.p238 td { padding: 0px 3px 0px 3px; } + +table.p238 td { padding: 0px 3px 0px 3px; } Create a sequence of numbers using the "Blum Blum Shub" pseudo-random number generator: @@ -17,18 +17,18 @@ sn+1 sn2 mod 20300713 -Concatenate these numbers  s0s1s2… to create a string w of infinite length. +Concatenate these numbers  s0s1s2… to create a string w of infinite length. Then, w = 14025256741014958470038053646… For a positive integer k, if no substring of w exists with a sum of digits equal to k, p(k) is defined to be zero. If at least one substring of w exists with a sum of digits equal to k, we define p(k) = z, where z is the starting position of the earliest such substring. For instance: -The substrings 1, 14, 1402, … -with respective sums of digits equal to 1, 5, 7, … +The substrings 1, 14, 1402, … +with respective sums of digits equal to 1, 5, 7, … start at position 1, hence p(1) = p(5) = p(7) = … = 1. -The substrings 4, 402, 4025, … -with respective sums of digits equal to 4, 6, 11, … +The substrings 4, 402, 4025, … +with respective sums of digits equal to 4, 6, 11, … start at position 2, hence p(4) = p(6) = p(11) = … = 2. -The substrings 02, 0252, … -with respective sums of digits equal to 2, 9, … +The substrings 02, 0252, … +with respective sums of digits equal to 2, 9, … start at position 3, hence p(2) = p(9) = … = 3. Note that substring 025 starting at position 3, has a sum of digits equal to 7, but there was an earlier substring (starting at position 1) with a sum of digits equal to 7, so p(7) = 1, not 3. We can verify that, for 0 k  103,  p(k) = 4742. diff --git a/project_euler/problems/239_twentytwo_foolish_primes.txt b/project_euler/problems/239_twentytwo_foolish_primes.txt index 5cbf017..0d2f692 100644 --- a/project_euler/problems/239_twentytwo_foolish_primes.txt +++ b/project_euler/problems/239_twentytwo_foolish_primes.txt @@ -3,7 +3,7 @@ http://projecteuler.net/problem=239 Twenty-two Foolish Primes A set of disks numbered 1 through 100 are placed in a line in random order. -What is the probability that we have a partial derangement such that exactly 22 prime number discs are found away from their natural positions? +What is the probability that we have a partial derangement such that exactly 22 prime number discs are found away from their natural positions? (Any number of non-prime disks may also be found in or out of their natural positions.) Give your answer rounded to 12 places behind the decimal point in the form 0.abcdefghijkl. diff --git a/project_euler/problems/240_top_dice.txt b/project_euler/problems/240_top_dice.txt index 3687c13..43ae801 100644 --- a/project_euler/problems/240_top_dice.txt +++ b/project_euler/problems/240_top_dice.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=240 Top Dice -There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can be rolled so that the top three sum to 15. Some examples are: - - -D1,D2,D3,D4,D5 = 4,3,6,3,5 - -D1,D2,D3,D4,D5 = 4,3,3,5,6 - -D1,D2,D3,D4,D5 = 3,3,3,6,6 - -D1,D2,D3,D4,D5 = 6,6,3,3,3 - +There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can be rolled so that the top three sum to 15. Some examples are: + + +D1,D2,D3,D4,D5 = 4,3,6,3,5 + +D1,D2,D3,D4,D5 = 4,3,3,5,6 + +D1,D2,D3,D4,D5 = 3,3,3,6,6 + +D1,D2,D3,D4,D5 = 6,6,3,3,3 + In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be rolled so that the top ten sum to 70? diff --git a/project_euler/problems/241_perfection_quotients.txt b/project_euler/problems/241_perfection_quotients.txt index 1996cfc..e4ee1ac 100644 --- a/project_euler/problems/241_perfection_quotients.txt +++ b/project_euler/problems/241_perfection_quotients.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=241 Perfection Quotients -For a positive integer n, let σ(n) be the sum of all divisors of n, so e.g. σ(6) = 1 + 2 + 3 + 6 = 12. +For a positive integer n, let σ(n) be the sum of all divisors of n, so e.g. σ(6) = 1 + 2 + 3 + 6 = 12. A perfect number, as you probably know, is a number with σ(n) = 2n. diff --git a/project_euler/problems/242_odd_triplets.txt b/project_euler/problems/242_odd_triplets.txt index 8e300e3..2479d17 100644 --- a/project_euler/problems/242_odd_triplets.txt +++ b/project_euler/problems/242_odd_triplets.txt @@ -3,9 +3,9 @@ http://projecteuler.net/problem=242 Odd Triplets Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element subsets with an odd sum of elements. For example, f(5,3) = 4, since the set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements, i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}. -When all three values n, k and f(n,k) are odd, we say that they make +When all three values n, k and f(n,k) are odd, we say that they make an odd-triplet [n,k,f(n,k)]. -There are exactly five odd-triplets with n  10, namely: +There are exactly five odd-triplets with n  10, namely: [1,1,f(1,1) = 1], [5,1,f(5,1) = 3], [5,5,f(5,5) = 1], [9,1,f(9,1) = 5] and [9,9,f(9,9) = 1]. How many odd-triplets are there with n  1012 ? diff --git a/project_euler/problems/244_sliders.txt b/project_euler/problems/244_sliders.txt index c4e6f98..401eb43 100644 --- a/project_euler/problems/244_sliders.txt +++ b/project_euler/problems/244_sliders.txt @@ -9,16 +9,16 @@ A move is denoted by the uppercase initial of the direction (Left, Right, Up, Do (S), (E) -For each path, its checksum is calculated by (pseudocode): - -checksum = 0 -checksum = (checksum 243 + m1) mod 100 000 007 -checksum = (checksum 243 + m2) mod 100 000 007 -   … +For each path, its checksum is calculated by (pseudocode): + +checksum = 0 +checksum = (checksum 243 + m1) mod 100 000 007 +checksum = (checksum 243 + m2) mod 100 000 007 +   … checksum = (checksum 243 + mn) mod 100 000 007 - -where mk is the ASCII value of the kth letter in the move sequence and the ASCII values for the moves are: - + +where mk is the ASCII value of the kth letter in the move sequence and the ASCII values for the moves are: + L76 @@ -28,7 +28,7 @@ D68 For the sequence LULUR given above, the checksum would be 19761398. -Now, starting from configuration (S), +Now, starting from configuration (S), find all shortest ways to reach configuration (T). diff --git a/project_euler/problems/245_coresilience.txt b/project_euler/problems/245_coresilience.txt index dd133aa..0edbb0d 100644 --- a/project_euler/problems/245_coresilience.txt +++ b/project_euler/problems/245_coresilience.txt @@ -26,6 +26,6 @@ The coresilience of a prime p is C(p) -Find the sum of all composite integers 1 n 21011, for which C(n) is a unit fraction. +Find the sum of all composite integers 1 n 21011, for which C(n) is a unit fraction. diff --git a/project_euler/problems/246_tangents_to_an_ellipse.txt b/project_euler/problems/246_tangents_to_an_ellipse.txt index 4e2883e..0534550 100644 --- a/project_euler/problems/246_tangents_to_an_ellipse.txt +++ b/project_euler/problems/246_tangents_to_an_ellipse.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=246 Tangents to an ellipse - -A definition for an ellipse is: -Given a circle c with centre M and radius r and a point G such that d(G,M)r, the locus of the points that are equidistant from c and G form an ellipse. - -The construction of the points of the ellipse is shown below. - +A definition for an ellipse is: +Given a circle c with centre M and radius r and a point G such that d(G,M)r, the locus of the points that are equidistant from c and G form an ellipse. - -Given are the points M(-2000,1500) and G(8000,1500). -Given is also the circle c with centre M and radius 15000. -The locus of the points that are equidistant from G and c form an ellipse e. -From a point P outside e the two tangents t1 and t2 to the ellipse are drawn. -Let the points where t1 and t2 touch the ellipse be R and S. +The construction of the points of the ellipse is shown below. - -For how many lattice points P is angle RPS greater than 45 degrees? + +Given are the points M(-2000,1500) and G(8000,1500). +Given is also the circle c with centre M and radius 15000. +The locus of the points that are equidistant from G and c form an ellipse e. +From a point P outside e the two tangents t1 and t2 to the ellipse are drawn. +Let the points where t1 and t2 touch the ellipse be R and S. + + + + +For how many lattice points P is angle RPS greater than 45 degrees? diff --git a/project_euler/problems/247_squares_under_a_hyperbola.txt b/project_euler/problems/247_squares_under_a_hyperbola.txt index 3cffb73..24b9eaa 100644 --- a/project_euler/problems/247_squares_under_a_hyperbola.txt +++ b/project_euler/problems/247_squares_under_a_hyperbola.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=247 Squares under a hyperbola -Consider the region constrained by 1 x and 0 y 1/x. - -Let S1 be the largest square that can fit under the curve. -Let S2 be the largest square that fits in the remaining area, and so on. -Let the index of Sn be the pair (left, below) indicating the number of squares to the left of Sn and the number of squares below Sn. +Consider the region constrained by 1 x and 0 y 1/x. +Let S1 be the largest square that can fit under the curve. +Let S2 be the largest square that fits in the remaining area, and so on. +Let the index of Sn be the pair (left, below) indicating the number of squares to the left of Sn and the number of squares below Sn. - -The diagram shows some such squares labelled by number. -S2 has one square to its left and none below, so the index of S2 is (1,0). -It can be seen that the index of S32 is (1,1) as is the index of S50. -50 is the largest n for which the index of Sn is (1,1). - -What is the largest n for which the index of Sn is (3,3)? + +The diagram shows some such squares labelled by number. +S2 has one square to its left and none below, so the index of S2 is (1,0). +It can be seen that the index of S32 is (1,1) as is the index of S50. +50 is the largest n for which the index of Sn is (1,1). + + +What is the largest n for which the index of Sn is (3,3)? diff --git a/project_euler/problems/249_prime_subset_sums.txt b/project_euler/problems/249_prime_subset_sums.txt index ee8916f..81c991d 100644 --- a/project_euler/problems/249_prime_subset_sums.txt +++ b/project_euler/problems/249_prime_subset_sums.txt @@ -3,6 +3,6 @@ http://projecteuler.net/problem=249 Prime Subset Sums Let S = {2, 3, 5, ..., 4999} be the set of prime numbers less than 5000. -Find the number of subsets of S, the sum of whose elements is a prime number. +Find the number of subsets of S, the sum of whose elements is a prime number. Enter the rightmost 16 digits as your answer. diff --git a/project_euler/problems/251_cardano_triplets.txt b/project_euler/problems/251_cardano_triplets.txt index fa22704..a3d1b82 100644 --- a/project_euler/problems/251_cardano_triplets.txt +++ b/project_euler/problems/251_cardano_triplets.txt @@ -2,18 +2,16 @@ http://projecteuler.net/problem=251 Cardano Triplets - + A triplet of positive integers (a,b,c) is called a Cardano Triplet if it satisfies the condition: - -For example, (2,1,5) is a Cardano Triplet. - -There exist 149 Cardano Triplets for which a+b+c 1000. +For example, (2,1,5) is a Cardano Triplet. + + +There exist 149 Cardano Triplets for which a+b+c 1000. + - -Find how many Cardano Triplets exist such that a+b+c 110,000,000. - - +Find how many Cardano Triplets exist such that a+b+c 110,000,000. diff --git a/project_euler/problems/252_convex_holes.txt b/project_euler/problems/252_convex_holes.txt index 663daaf..aac0162 100644 --- a/project_euler/problems/252_convex_holes.txt +++ b/project_euler/problems/252_convex_holes.txt @@ -2,22 +2,22 @@ http://projecteuler.net/problem=252 Convex Holes - -Given a set of points on a plane, we define a convex hole to be a convex polygon having as vertices any of the given points and not containing any of the given points in its interior (in addition to the vertices, other given points may lie on the perimeter of the polygon). - -As an example, the image below shows a set of twenty points and a few such convex holes. -The convex hole shown as a red heptagon has an area equal to 1049694.5 square units, which is the highest possible area for a convex hole on the given set of points. +Given a set of points on a plane, we define a convex hole to be a convex polygon having as vertices any of the given points and not containing any of the given points in its interior (in addition to the vertices, other given points may lie on the perimeter of the polygon). +As an example, the image below shows a set of twenty points and a few such convex holes. +The convex hole shown as a red heptagon has an area equal to 1049694.5 square units, which is the highest possible area for a convex hole on the given set of points. - -table.p252 td { - padding: 0px 3px 0px 3px; - vertical-align: bottom; - text-align: left; -} + + + +table.p252 td { + padding: 0px 3px 0px 3px; + vertical-align: bottom; + text-align: left; +} For our example, we used the first 20 points (T2k1, T2k), for k = 1,2,…,20, produced with the pseudo-random number generator: @@ -36,9 +36,9 @@ Tn -i.e. (527, 144), (488, 732), (454, 947), … +i.e. (527, 144), (488, 732), (454, 947), … + - -What is the maximum area for a convex hole on the set containing the first 500 points in the pseudo-random sequence? Specify your answer including one digit after the decimal point. +What is the maximum area for a convex hole on the set containing the first 500 points in the pseudo-random sequence? Specify your answer including one digit after the decimal point. diff --git a/project_euler/problems/253_tidying_up.txt b/project_euler/problems/253_tidying_up.txt index 94362f0..e77383e 100644 --- a/project_euler/problems/253_tidying_up.txt +++ b/project_euler/problems/253_tidying_up.txt @@ -19,7 +19,7 @@ Segments So Far 354 …… -Let M be the maximum number of segments encountered during a random tidy-up of the caterpillar. +Let M be the maximum number of segments encountered during a random tidy-up of the caterpillar. For a caterpillar of ten pieces, the number of possibilities for each M is diff --git a/project_euler/problems/255_rounded_square_roots.txt b/project_euler/problems/255_rounded_square_roots.txt index 37eb34a..0f704a3 100644 --- a/project_euler/problems/255_rounded_square_roots.txt +++ b/project_euler/problems/255_rounded_square_roots.txt @@ -4,28 +4,28 @@ Rounded Square Roots We define the rounded-square-root of a positive integer n as the square root of n rounded to the nearest integer. The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of n: -Let d be the number of digits of the number n. -If d is odd, set x0 = 210(d-1)⁄2. -If d is even, set x0 = 710(d-2)⁄2. +Let d be the number of digits of the number n. +If d is odd, set x0 = 210(d-1)⁄2. +If d is even, set x0 = 710(d-2)⁄2. Repeat: -until xk+1 = xk. +until xk+1 = xk. As an example, let us find the rounded-square-root of n = 4321. n has 4 digits, so x0 = 710(4-2)⁄2 = 70. - -Since x2 = x1, we stop here. -So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…). -The number of iterations required when using this method is surprisingly low. -For example, we can find the rounded-square-root of a 5-digit integer (10,000 n 99,999) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places). +Since x2 = x1, we stop here. +So, after just two iterations, we have found that the rounded-square-root of 4321 is 66 (the actual square root is 65.7343137…). -Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (1013 n 1014)? -Give your answer rounded to 10 decimal places. +The number of iterations required when using this method is surprisingly low. +For example, we can find the rounded-square-root of a 5-digit integer (10,000 n 99,999) with an average of 3.2102888889 iterations (the average value was rounded to 10 decimal places). -Note: The symbols x and x represent the floor function and ceiling function respectively. +Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a 14-digit number (1013 n 1014)? +Give your answer rounded to 10 decimal places. + +Note: The symbols x and x represent the floor function and ceiling function respectively. diff --git a/project_euler/problems/256_tatamifree_rooms.txt b/project_euler/problems/256_tatamifree_rooms.txt index 0a7aeb1..988cfef 100644 --- a/project_euler/problems/256_tatamifree_rooms.txt +++ b/project_euler/problems/256_tatamifree_rooms.txt @@ -5,21 +5,21 @@ Tatami-Free Rooms Tatami are rectangular mats, used to completely cover the floor of a room, without overlap. Assuming that the only type of available tatami has dimensions 12, there are obviously some limitations for the shape and size of the rooms that can be covered. -For this problem, we consider only rectangular rooms with integer dimensions a, b and even size s = a·b. +For this problem, we consider only rectangular rooms with integer dimensions a, b and even size s = a·b. We use the term 'size' to denote the floor surface area of the room, and — without loss of generality — we add the condition a b. -There is one rule to follow when laying out tatami: there must be no points where corners of four different mats meet. +There is one rule to follow when laying out tatami: there must be no points where corners of four different mats meet. For example, consider the two arrangements below for a 44 room: The arrangement on the left is acceptable, whereas the one on the right is not: a red "X" in the middle, marks the point where four tatami meet. -Because of this rule, certain even-sized rooms cannot be covered with tatami: we call them tatami-free rooms. +Because of this rule, certain even-sized rooms cannot be covered with tatami: we call them tatami-free rooms. Further, we define T(s) as the number of tatami-free rooms of size s. -The smallest tatami-free room has size s = 70 and dimensions 710. -All the other rooms of size s = 70 can be covered with tatami; they are: 170, 235 and 514. +The smallest tatami-free room has size s = 70 and dimensions 710. +All the other rooms of size s = 70 can be covered with tatami; they are: 170, 235 and 514. Hence, T(70) = 1. -Similarly, we can verify that T(1320) = 5 because there are exactly 5 tatami-free rooms of size s = 1320: -2066, 2260, 2455, 3044 and 3340. +Similarly, we can verify that T(1320) = 5 because there are exactly 5 tatami-free rooms of size s = 1320: +2066, 2260, 2455, 3044 and 3340. In fact, s = 1320 is the smallest room-size s for which T(s) = 5. Find the smallest room-size s for which T(s) = 200. diff --git a/project_euler/problems/257_angular_bisectors.txt b/project_euler/problems/257_angular_bisectors.txt index 917a1ed..6c41c11 100644 --- a/project_euler/problems/257_angular_bisectors.txt +++ b/project_euler/problems/257_angular_bisectors.txt @@ -3,20 +3,20 @@ http://projecteuler.net/problem=257 Angular Bisectors -Given is an integer sided triangle ABC with sides a b c. -(AB = c, BC = a and AC = b). -The angular bisectors of the triangle intersect the sides at points E, F and G (see picture below). +Given is an integer sided triangle ABC with sides a b c. +(AB = c, BC = a and AC = b). +The angular bisectors of the triangle intersect the sides at points E, F and G (see picture below). - -The segments EF, EG and FG partition the triangle ABC into four smaller triangles: AEG, BFE, CGF and EFG. -It can be proven that for each of these four triangles the ratio area(ABC)/area(subtriangle) is rational. -However, there exist triangles for which some or all of these ratios are integral. - -How many triangles ABC with perimeter100,000,000 exist so that the ratio area(ABC)/area(AEG) is integral? +The segments EF, EG and FG partition the triangle ABC into four smaller triangles: AEG, BFE, CGF and EFG. +It can be proven that for each of these four triangles the ratio area(ABC)/area(subtriangle) is rational. +However, there exist triangles for which some or all of these ratios are integral. + + +How many triangles ABC with perimeter100,000,000 exist so that the ratio area(ABC)/area(AEG) is integral? diff --git a/project_euler/problems/258_a_lagged_fibonacci_sequence.txt b/project_euler/problems/258_a_lagged_fibonacci_sequence.txt index 76cde6b..e5c645d 100644 --- a/project_euler/problems/258_a_lagged_fibonacci_sequence.txt +++ b/project_euler/problems/258_a_lagged_fibonacci_sequence.txt @@ -5,7 +5,7 @@ A lagged Fibonacci sequence A sequence is defined as: gk = 1, for 0 k 1999 -gk = gk-2000 + gk-1999, for k 2000. +gk = gk-2000 + gk-1999, for k 2000. -Find gk mod 20092010 for k = 1018. +Find gk mod 20092010 for k = 1018. diff --git a/project_euler/problems/260_stone_game.txt b/project_euler/problems/260_stone_game.txt index 8eda2ca..68c1731 100644 --- a/project_euler/problems/260_stone_game.txt +++ b/project_euler/problems/260_stone_game.txt @@ -2,20 +2,20 @@ http://projecteuler.net/problem=260 Stone Game -A game is played with three piles of stones and two players. +A game is played with three piles of stones and two players. At her turn, a player removes one or more stones from the piles. However, if she takes stones from more than one pile, she must remove the same number of stones from each of the selected piles. -In other words, the player chooses some N>0 and removes: +In other words, the player chooses some N>0 and removes: N stones from any single pile; or N stones from each of any two piles (2N total); or -N stones from each of the three piles (3N total). -The player taking the last stone(s) wins the game. - -A winning configuration is one where the first player can force a win. +N stones from each of the three piles (3N total). +The player taking the last stone(s) wins the game. + +A winning configuration is one where the first player can force a win. For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations because the first player can immediately remove all stones. -A losing configuration is one where the second player can force a win, no matter what the first player does. +A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (0,1,2) and (1,3,3) are losing configurations: any legal move leaves a winning configuration for the second player. -Consider all losing configurations (xi,yi,zi) where xi yi zi 100. +Consider all losing configurations (xi,yi,zi) where xi yi zi 100. We can verify that Σ(xi+yi+zi) = 173895 for these. -Find Σ(xi+yi+zi) where (xi,yi,zi) ranges over the losing configurations +Find Σ(xi+yi+zi) where (xi,yi,zi) ranges over the losing configurations with xi yi zi 1000. diff --git a/project_euler/problems/261_pivotal_square_sums.txt b/project_euler/problems/261_pivotal_square_sums.txt index f22d2ae..c97943d 100644 --- a/project_euler/problems/261_pivotal_square_sums.txt +++ b/project_euler/problems/261_pivotal_square_sums.txt @@ -3,10 +3,10 @@ http://projecteuler.net/problem=261 Pivotal Square Sums Let us call a positive integer k a square-pivot, if there is a pair of integers m 0 and n k, such that the sum of the (m+1) consecutive squares up to k equals the sum of the m consecutive squares from (n+1) on: - + (k-m)2 + ... + k2 = (n+1)2 + ... + (n+m)2. -Some small square-pivots are -4: 32 + 42 +Some small square-pivots are +4: 32 + 42 = 52 21: 202 + 212 = 292 24: 212 + 222 + 232 + 242 = 252 + 262 + 272 diff --git a/project_euler/problems/262_mountain_range.txt b/project_euler/problems/262_mountain_range.txt index 93cc972..621659e 100644 --- a/project_euler/problems/262_mountain_range.txt +++ b/project_euler/problems/262_mountain_range.txt @@ -8,9 +8,9 @@ The following equation represents the continuous topography of a mountainous reg A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving the area given by 0 ≤ x, y ≤ 1600. Because of the intervening mountains, it first rises straight up to a point A', having elevation f. Then, while remaining at the same elevation f, it flies around any obstacles until it arrives at a point B' directly above B. -First, determine fmin which is the minimum constant elevation allowing such a trip from A to B, while remaining in the specified area. +First, determine fmin which is the minimum constant elevation allowing such a trip from A to B, while remaining in the specified area. Then, find the length of the shortest path between A' and B', while flying at that constant elevation fmin. Give that length as your answer, rounded to three decimal places. -Note: For convenience, the elevation function shown above is repeated below, in a form suitable for most programming languages: +Note: For convenience, the elevation function shown above is repeated below, in a form suitable for most programming languages: h=( 5000-0.005*(x*x+y*y+x*y)+12.5*(x+y) ) * exp( -abs(0.000001*(x*x+y*y)-0.0015*(x+y)+0.7) ) diff --git a/project_euler/problems/263_an_engineers_dream_come_true.txt b/project_euler/problems/263_an_engineers_dream_come_true.txt index 5a7ccb1..d281eb2 100644 --- a/project_euler/problems/263_an_engineers_dream_come_true.txt +++ b/project_euler/problems/263_an_engineers_dream_come_true.txt @@ -2,27 +2,27 @@ http://projecteuler.net/problem=263 An engineers' dream come true - -Consider the number 6. The divisors of 6 are: 1,2,3 and 6. -Every number from 1 up to and including 6 can be written as a sum of distinct divisors of 6: -1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6. -A number n is called a practical number if every number from 1 up to and including n can be expressed as a sum of distinct divisors of n. - - -A pair of consecutive prime numbers with a difference of six is called a sexy pair (since "sex" is the Latin word for "six"). The first sexy pair is (23, 29). - - -We may occasionally find a triple-pair, which means three consecutive sexy prime pairs, such that the second member of each pair is the first member of the next pair. - - -We shall call a number n such that : - -(n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and -the numbers n-8, n-4, n, n+4 and n+8 are all practical, - -an engineers’ paradise. - - -Find the sum of the first four engineers’ paradises. + +Consider the number 6. The divisors of 6 are: 1,2,3 and 6. +Every number from 1 up to and including 6 can be written as a sum of distinct divisors of 6: +1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6. +A number n is called a practical number if every number from 1 up to and including n can be expressed as a sum of distinct divisors of n. + + +A pair of consecutive prime numbers with a difference of six is called a sexy pair (since "sex" is the Latin word for "six"). The first sexy pair is (23, 29). + + +We may occasionally find a triple-pair, which means three consecutive sexy prime pairs, such that the second member of each pair is the first member of the next pair. + + +We shall call a number n such that : + +(n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and +the numbers n-8, n-4, n, n+4 and n+8 are all practical, + +an engineers’ paradise. + + +Find the sum of the first four engineers’ paradises. diff --git a/project_euler/problems/264_triangle_centres.txt b/project_euler/problems/264_triangle_centres.txt index 626aeaf..4c1da45 100644 --- a/project_euler/problems/264_triangle_centres.txt +++ b/project_euler/problems/264_triangle_centres.txt @@ -2,29 +2,29 @@ http://projecteuler.net/problem=264 Triangle Centres -Consider all the triangles having: +Consider all the triangles having: All their vertices on lattice points. Circumcentre at the origin O. Orthocentre at the point H(5, 0). -There are nine such triangles having a perimeter 50. +There are nine such triangles having a perimeter 50. Listed and shown in ascending order of their perimeter, they are: -A(-4, 3), B(5, 0), C(4, -3) -A(4, 3), B(5, 0), C(-4, -3) -A(-3, 4), B(5, 0), C(3, -4) -A(3, 4), B(5, 0), C(-3, -4) -A(0, 5), B(5, 0), C(0, -5) -A(1, 8), B(8, -1), C(-4, -7) -A(8, 1), B(1, -8), C(-4, 7) -A(2, 9), B(9, -2), C(-6, -7) +A(-4, 3), B(5, 0), C(4, -3) +A(4, 3), B(5, 0), C(-4, -3) +A(-3, 4), B(5, 0), C(3, -4) +A(3, 4), B(5, 0), C(-3, -4) +A(0, 5), B(5, 0), C(0, -5) +A(1, 8), B(8, -1), C(-4, -7) +A(8, 1), B(1, -8), C(-4, 7) +A(2, 9), B(9, -2), C(-6, -7) A(9, 2), B(2, -9), C(-6, 7) The sum of their perimeters, rounded to four decimal places, is 291.0089. -Find all such triangles with a perimeter 105. +Find all such triangles with a perimeter 105. Enter as your answer the sum of their perimeters rounded to four decimal places. diff --git a/project_euler/problems/265_binary_circles.txt b/project_euler/problems/265_binary_circles.txt index b372b50..ca77aa4 100644 --- a/project_euler/problems/265_binary_circles.txt +++ b/project_euler/problems/265_binary_circles.txt @@ -6,7 +6,7 @@ Binary Circles For N=3, two such circular arrangements are possible, ignoring rotations: For the first arrangement, the 3-digit subsequences, in clockwise order, are: 000, 001, 010, 101, 011, 111, 110 and 100. -Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29: +Each circular arrangement can be encoded as a number by concatenating the binary digits starting with the subsequence of all zeros as the most significant bits and proceeding clockwise. The two arrangements for N=3 are thus represented as 23 and 29: 00010111 2 = 23 00011101 2 = 29 Calling S(N) the sum of the unique numeric representations, we can see that S(3) = 23 + 29 = 52. diff --git a/project_euler/problems/266_pseudo_square_root.txt b/project_euler/problems/266_pseudo_square_root.txt index a1fbc9f..95b516b 100644 --- a/project_euler/problems/266_pseudo_square_root.txt +++ b/project_euler/problems/266_pseudo_square_root.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=266 Pseudo Square Root - -The divisors of 12 are: 1,2,3,4,6 and 12. -The largest divisor of 12 that does not exceed the square root of 12 is 3. -We shall call the largest divisor of an integer n that does not exceed the square root of n the pseudo square root (PSR) of n. -It can be seen that PSR(3102)=47. - - -Let p be the product of the primes below 190. -Find PSR(p) mod 1016. + +The divisors of 12 are: 1,2,3,4,6 and 12. +The largest divisor of 12 that does not exceed the square root of 12 is 3. +We shall call the largest divisor of an integer n that does not exceed the square root of n the pseudo square root (PSR) of n. +It can be seen that PSR(3102)=47. + + +Let p be the product of the primes below 190. +Find PSR(p) mod 1016. diff --git a/project_euler/problems/269_polynomials_with_at_least_one_integer_root.txt b/project_euler/problems/269_polynomials_with_at_least_one_integer_root.txt index cb80de4..4fe16f0 100644 --- a/project_euler/problems/269_polynomials_with_at_least_one_integer_root.txt +++ b/project_euler/problems/269_polynomials_with_at_least_one_integer_root.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=269 Polynomials with at least one integer root -A root or zero of a polynomial P(x) is a solution to the equation P(x) = 0. -Define Pn as the polynomial whose coefficients are the digits of n. +A root or zero of a polynomial P(x) is a solution to the equation P(x) = 0. +Define Pn as the polynomial whose coefficients are the digits of n. For example, P5703(x) = 5x3 + 7x2 + 3. We can see that: Pn(0) is the last digit of n, diff --git a/project_euler/problems/270_cutting_squares.txt b/project_euler/problems/270_cutting_squares.txt index 0bab1d3..1068e69 100644 --- a/project_euler/problems/270_cutting_squares.txt +++ b/project_euler/problems/270_cutting_squares.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=270 Cutting Squares -A square piece of paper with integer dimensions NN is placed with a corner at the origin and two of its sides along the x- and y-axes. Then, we cut it up respecting the following rules: +A square piece of paper with integer dimensions NN is placed with a corner at the origin and two of its sides along the x- and y-axes. Then, we cut it up respecting the following rules: We only make straight cuts between two points lying on different sides of the square, and having integer coordinates. Two cuts cannot cross, but several cuts can meet at the same border point. diff --git a/project_euler/problems/271_modular_cubes_part_1.txt b/project_euler/problems/271_modular_cubes_part_1.txt index 8af4708..c3ccb87 100644 --- a/project_euler/problems/271_modular_cubes_part_1.txt +++ b/project_euler/problems/271_modular_cubes_part_1.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=271 Modular Cubes, part 1 - -For a positive number n, define S(n) as the sum of the integers x, for which 1xn and x31 mod n. - -When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. +For a positive number n, define S(n) as the sum of the integers x, for which 1xn and x31 mod n. + + +When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. Thus, S(91)=9+16+22+29+53+74+79+81=363. - -Find S(13082761331670030). + +Find S(13082761331670030). diff --git a/project_euler/problems/272_modular_cubes_part_2.txt b/project_euler/problems/272_modular_cubes_part_2.txt index 08d3f43..84e2543 100644 --- a/project_euler/problems/272_modular_cubes_part_2.txt +++ b/project_euler/problems/272_modular_cubes_part_2.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=272 Modular Cubes, part 2 - -For a positive number n, define C(n) as the number of the integers x, for which 1xn and x31 mod n. - -When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. +For a positive number n, define C(n) as the number of the integers x, for which 1xn and x31 mod n. + + +When n=91, there are 8 possible values for x, namely : 9, 16, 22, 29, 53, 74, 79, 81. Thus, C(91)=8. - -Find the sum of the positive numbers n1011 for which C(n)=242. + +Find the sum of the positive numbers n1011 for which C(n)=242. diff --git a/project_euler/problems/275_balanced_sculptures.txt b/project_euler/problems/275_balanced_sculptures.txt index 705950f..0b253b5 100644 --- a/project_euler/problems/275_balanced_sculptures.txt +++ b/project_euler/problems/275_balanced_sculptures.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=275 Balanced Sculptures -Let us define a balanced sculpture of order n as follows: +Let us define a balanced sculpture of order n as follows: A polyomino made up of n+1 tiles known as the blocks (n tiles) and the plinth (remaining tile); the plinth has its centre at position (x = 0, y = 0); diff --git a/project_euler/problems/276_primitive_triangles.txt b/project_euler/problems/276_primitive_triangles.txt index 686962b..24f6694 100644 --- a/project_euler/problems/276_primitive_triangles.txt +++ b/project_euler/problems/276_primitive_triangles.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=276 Primitive Triangles -Consider the triangles with integer sides a, b and c with a b c. -An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. -How many primitive integer sided triangles exist with a perimeter not exceeding 10 000 000? +Consider the triangles with integer sides a, b and c with a b c. +An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. +How many primitive integer sided triangles exist with a perimeter not exceeding 10 000 000? diff --git a/project_euler/problems/277_a_modified_collatz_sequence.txt b/project_euler/problems/277_a_modified_collatz_sequence.txt index dbb5216..f996819 100644 --- a/project_euler/problems/277_a_modified_collatz_sequence.txt +++ b/project_euler/problems/277_a_modified_collatz_sequence.txt @@ -2,29 +2,29 @@ http://projecteuler.net/problem=277 A Modified Collatz sequence - + A modified Collatz sequence of integers is obtained from a starting value a1 in the following way: an+1 = an/3 if an is divisible by 3. We shall denote this as a large downward step, "D". -an+1 = (4an + 2)/3 if an divided by 3 gives a remainder of 1. We shall denote this as an upward step, "U". +an+1 = (4an + 2)/3 if an divided by 3 gives a remainder of 1. We shall denote this as an upward step, "U". + + +an+1 = (2an - 1)/3 if an divided by 3 gives a remainder of 2. We shall denote this as a small downward step, "d". + + +The sequence terminates when some an = 1. -an+1 = (2an - 1)/3 if an divided by 3 gives a remainder of 2. We shall denote this as a small downward step, "d". +Given any integer, we can list out the sequence of steps. +For instance if a1=231, then the sequence {an}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps "DdDddUUdDD". - -The sequence terminates when some an = 1. - -Given any integer, we can list out the sequence of steps. -For instance if a1=231, then the sequence {an}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps "DdDddUUdDD". +Of course, there are other sequences that begin with that same sequence "DdDddUUdDD....". +For instance, if a1=1004064, then the sequence is DdDddUUdDDDdUDUUUdDdUUDDDUdDD. +In fact, 1004064 is the smallest possible a1 106 that begins with the sequence DdDddUUdDD. - -Of course, there are other sequences that begin with that same sequence "DdDddUUdDD....". -For instance, if a1=1004064, then the sequence is DdDddUUdDDDdUDUUUdDdUUDDDUdDD. -In fact, 1004064 is the smallest possible a1 106 that begins with the sequence DdDddUUdDD. - -What is the smallest a1 1015 that begins with the sequence "UDDDUdddDDUDDddDdDddDDUDDdUUDd"? +What is the smallest a1 1015 that begins with the sequence "UDDDUdddDDUDDddDdDddDDUDDdUUDd"? diff --git a/project_euler/problems/278_linear_combinations_of_semiprimes.txt b/project_euler/problems/278_linear_combinations_of_semiprimes.txt index 212d050..2666fb7 100644 --- a/project_euler/problems/278_linear_combinations_of_semiprimes.txt +++ b/project_euler/problems/278_linear_combinations_of_semiprimes.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=278 Linear Combinations of Semiprimes - -Given the values of integers 1 a1 a2 ... an, consider the linear combination q1a1 + q2a2 + ... + qnan = b, using only integer values qk 0. - - -Note that for a given set of ak, it may be that not all values of b are possible. -For instance, if a1 = 5 and a2 = 7, there are no q1 0 and q2 0 such that b could be -1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23. - -In fact, 23 is the largest impossible value of b for a1 = 5 and a2 = 7. We therefore call f(5, 7) = 23. Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195. - - -Find f(p*q,p*r,q*r), where p, q and r are prime numbers and p < q r 5000. - + +Given the values of integers 1 a1 a2 ... an, consider the linear combination q1a1 + q2a2 + ... + qnan = b, using only integer values qk 0. + + +Note that for a given set of ak, it may be that not all values of b are possible. +For instance, if a1 = 5 and a2 = 7, there are no q1 0 and q2 0 such that b could be +1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23. + +In fact, 23 is the largest impossible value of b for a1 = 5 and a2 = 7. We therefore call f(5, 7) = 23. Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195. + + +Find f(p*q,p*r,q*r), where p, q and r are prime numbers and p < q r 5000. + diff --git a/project_euler/problems/279_triangles_with_integral_sides_and_an_integral_angle.txt b/project_euler/problems/279_triangles_with_integral_sides_and_an_integral_angle.txt index 98a1d32..7eef4a3 100644 --- a/project_euler/problems/279_triangles_with_integral_sides_and_an_integral_angle.txt +++ b/project_euler/problems/279_triangles_with_integral_sides_and_an_integral_angle.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=279 Triangles with integral sides and an integral angle - -How many triangles are there with integral sides, at least one integral angle (measured in degrees), and a perimeter that does not exceed 108? + +How many triangles are there with integral sides, at least one integral angle (measured in degrees), and a perimeter that does not exceed 108? diff --git a/project_euler/problems/280_ant_and_seeds.txt b/project_euler/problems/280_ant_and_seeds.txt index 7d73716..13b3f8d 100644 --- a/project_euler/problems/280_ant_and_seeds.txt +++ b/project_euler/problems/280_ant_and_seeds.txt @@ -4,6 +4,6 @@ Ant and seeds A laborious ant walks randomly on a 5x5 grid. The walk starts from the central square. At each step, the ant moves to an adjacent square at random, without leaving the grid; thus there are 2, 3 or 4 possible moves at each step depending on the ant's position. At the start of the walk, a seed is placed on each square of the lower row. When the ant isn't carrying a seed and reaches a square of the lower row containing a seed, it will start to carry the seed. The ant will drop the seed on the first empty square of the upper row it eventually reaches. -What's the expected number of steps until all seeds have been dropped in the top row? +What's the expected number of steps until all seeds have been dropped in the top row? Give your answer rounded to 6 decimal places. diff --git a/project_euler/problems/282_the_ackermann_function.txt b/project_euler/problems/282_the_ackermann_function.txt index 1ed57d1..134568b 100644 --- a/project_euler/problems/282_the_ackermann_function.txt +++ b/project_euler/problems/282_the_ackermann_function.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=282 The Ackermann function - -For non-negative integers m, n, the Ackermann function A(m, n) is defined as follows: +For non-negative integers m, n, the Ackermann function A(m, n) is defined as follows: + + + +For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125. - -For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125. - Find A(n, n) and give your answer mod 148. diff --git a/project_euler/problems/283_integer_sided_triangles_for_which_the_areaperimeter_ratio_is_integral.txt b/project_euler/problems/283_integer_sided_triangles_for_which_the_areaperimeter_ratio_is_integral.txt index b4120c4..613d3ec 100644 --- a/project_euler/problems/283_integer_sided_triangles_for_which_the_areaperimeter_ratio_is_integral.txt +++ b/project_euler/problems/283_integer_sided_triangles_for_which_the_areaperimeter_ratio_is_integral.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=283 Integer sided triangles for which the area/perimeter ratio is integral - -Consider the triangle with sides 6, 8 and 10. It can be seen that the perimeter and the area are both equal to 24. -So the area/perimeter ratio is equal to 1. -Consider also the triangle with sides 13, 14 and 15. The perimeter equals 42 while the area is equal to 84. -So for this triangle the area/perimeter ratio is equal to 2. - - -Find the sum of the perimeters of all integer sided triangles for which the area/perimeter ratios are equal to positive integers not exceeding 1000. + +Consider the triangle with sides 6, 8 and 10. It can be seen that the perimeter and the area are both equal to 24. +So the area/perimeter ratio is equal to 1. +Consider also the triangle with sides 13, 14 and 15. The perimeter equals 42 while the area is equal to 84. +So for this triangle the area/perimeter ratio is equal to 2. + + +Find the sum of the perimeters of all integer sided triangles for which the area/perimeter ratios are equal to positive integers not exceeding 1000. diff --git a/project_euler/problems/284_steady_squares.txt b/project_euler/problems/284_steady_squares.txt index c11eb99..a9bccf1 100644 --- a/project_euler/problems/284_steady_squares.txt +++ b/project_euler/problems/284_steady_squares.txt @@ -5,6 +5,6 @@ Steady Squares The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 3762 = 141376. Let's call a number with this property a steady square. Steady squares can also be observed in other numbering systems. In the base 14 numbering system, the 3-digit number c37 is also a steady square: c372 = aa0c37, and the sum of its digits is c+3+7=18 in the same numbering system. The letters a, b, c and d are used for the 10, 11, 12 and 13 digits respectively, in a manner similar to the hexadecimal numbering system. For 1 n 9, the sum of the digits of all the n-digit steady squares in the base 14 numbering system is 2d8 (582 decimal). Steady squares with leading 0's are not allowed. -Find the sum of the digits of all the n-digit steady squares in the base 14 numbering system for +Find the sum of the digits of all the n-digit steady squares in the base 14 numbering system for 1 n 10000 (decimal) and give your answer in the base 14 system using lower case letters where necessary. diff --git a/project_euler/problems/285_pythagorean_odds.txt b/project_euler/problems/285_pythagorean_odds.txt index 45bb45a..1ccd2fb 100644 --- a/project_euler/problems/285_pythagorean_odds.txt +++ b/project_euler/problems/285_pythagorean_odds.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=285 Pythagorean odds -Albert chooses a positive integer k, then two real numbers a, b are randomly chosen in the interval [0,1] with uniform distribution. +Albert chooses a positive integer k, then two real numbers a, b are randomly chosen in the interval [0,1] with uniform distribution. The square root of the sum (k·a+1)2 + (k·b+1)2 is then computed and rounded to the nearest integer. If the result is equal to k, he scores k points; otherwise he scores nothing. -For example, if k = 6, a = 0.2 and b = 0.85, then (k·a+1)2 + (k·b+1)2 = 42.05. -The square root of 42.05 is 6.484... and when rounded to the nearest integer, it becomes 6. +For example, if k = 6, a = 0.2 and b = 0.85, then (k·a+1)2 + (k·b+1)2 = 42.05. +The square root of 42.05 is 6.484... and when rounded to the nearest integer, it becomes 6. This is equal to k, so he scores 6 points. It can be shown that if he plays 10 turns with k = 1, k = 2, ..., k = 10, the expected value of his total score, rounded to five decimal places, is 10.20914. If he plays 105 turns with k = 1, k = 2, k = 3, ..., k = 105, what is the expected value of his total score, rounded to five decimal places? diff --git a/project_euler/problems/287_quadtree_encoding_a_simple_compression_algorithm.txt b/project_euler/problems/287_quadtree_encoding_a_simple_compression_algorithm.txt index 4861f91..ad73731 100644 --- a/project_euler/problems/287_quadtree_encoding_a_simple_compression_algorithm.txt +++ b/project_euler/problems/287_quadtree_encoding_a_simple_compression_algorithm.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=287 Quadtree encoding (a simple compression algorithm) -The quadtree encoding allows us to describe a 2N2N black and white image as a sequence of bits (0 and 1). Those sequences are to be read from left to right like this: +The quadtree encoding allows us to describe a 2N2N black and white image as a sequence of bits (0 and 1). Those sequences are to be read from left to right like this: the first bit deals with the complete 2N2N region; -"0" denotes a split: -the current 2n2n region is divided into 4 sub-regions of dimension 2n-12n-1, +"0" denotes a split: +the current 2n2n region is divided into 4 sub-regions of dimension 2n-12n-1, the next bits contains the description of the top left, top right, bottom left and bottom right sub-regions - in that order; "10" indicates that the current region contains only black pixels; "11" indicates that the current region contains only white pixels. Consider the following 44 image (colored marks denote places where a split can occur): -This image can be described by several sequences, for example : -"001010101001011111011010101010", of length 30, or +This image can be described by several sequences, for example : +"001010101001011111011010101010", of length 30, or "0100101111101110", of length 16, which is the minimal sequence for this image. -For a positive integer N, define DN as the 2N2N image with the following coloring scheme: +For a positive integer N, define DN as the 2N2N image with the following coloring scheme: the pixel with coordinates x = 0, y = 0 corresponds to the bottom left pixel, if (x - 2N-1)2 + (y - 2N-1)2  22N-2 then the pixel is black, otherwise the pixel is white. diff --git a/project_euler/problems/288_an_enormous_factorial.txt b/project_euler/problems/288_an_enormous_factorial.txt index 9a55808..a3985ad 100644 --- a/project_euler/problems/288_an_enormous_factorial.txt +++ b/project_euler/problems/288_an_enormous_factorial.txt @@ -2,22 +2,22 @@ http://projecteuler.net/problem=288 An enormous factorial - -For any prime p the number N(p,q) is defined by + +For any prime p the number N(p,q) is defined by N(p,q) = n=0 to q Tn*pn with Tn generated by the following random number generator: - -S0 = 290797 -Sn+1 = Sn2 mod 50515093 + +S0 = 290797 +Sn+1 = Sn2 mod 50515093 Tn = Sn mod p - -Let Nfac(p,q) be the factorial of N(p,q). -Let NF(p,q) be the number of factors p in Nfac(p,q). - -You are given that NF(3,10000) mod 320=624955285. +Let Nfac(p,q) be the factorial of N(p,q). +Let NF(p,q) be the number of factors p in Nfac(p,q). + + +You are given that NF(3,10000) mod 320=624955285. + - Find NF(61,107) mod 6110 diff --git a/project_euler/problems/289_eulerian_cycles.txt b/project_euler/problems/289_eulerian_cycles.txt index 4510ec3..68584f4 100644 --- a/project_euler/problems/289_eulerian_cycles.txt +++ b/project_euler/problems/289_eulerian_cycles.txt @@ -3,14 +3,14 @@ http://projecteuler.net/problem=289 Eulerian Cycles Let C(x,y) be a circle passing through the points (x, y), (x, y+1), (x+1, y) and (x+1, y+1). -For positive integers m and n, let E(m,n) be a configuration which consists of the m·n circles: +For positive integers m and n, let E(m,n) be a configuration which consists of the m·n circles: { C(x,y): 0 ≤ x m, 0 ≤ y n, x and y are integers } -An Eulerian cycle on E(m,n) is a closed path that passes through each arc exactly once. -Many such paths are possible on E(m,n), but we are only interested in those which are not self-crossing: +An Eulerian cycle on E(m,n) is a closed path that passes through each arc exactly once. +Many such paths are possible on E(m,n), but we are only interested in those which are not self-crossing: A non-crossing path just touches itself at lattice points, but it never crosses itself. The image below shows E(3,3) and an example of an Eulerian non-crossing path. -Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n). +Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n). For example, L(1,2) = 2, L(2,2) = 37 and L(3,3) = 104290. Find L(6,10) mod 1010. diff --git a/project_euler/problems/291_panaitopol_primes.txt b/project_euler/problems/291_panaitopol_primes.txt index 52fc71b..b4c1eb0 100644 --- a/project_euler/problems/291_panaitopol_primes.txt +++ b/project_euler/problems/291_panaitopol_primes.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=291 Panaitopol Primes - -A prime number p is called a Panaitopol prime if for some positive integers x and y. - -Find how many Panaitopol primes are less than 51015. +A prime number p is called a Panaitopol prime if for some positive integers x and y. + + +Find how many Panaitopol primes are less than 51015. diff --git a/project_euler/problems/292_pythagorean_polygons.txt b/project_euler/problems/292_pythagorean_polygons.txt index 4127187..0d6ed72 100644 --- a/project_euler/problems/292_pythagorean_polygons.txt +++ b/project_euler/problems/292_pythagorean_polygons.txt @@ -8,8 +8,8 @@ there are at least three vertices, no three vertices are aligned, each vertex has integer coordinates, each edge has integer length. -For a given integer n, define P(n) as the number of distinct pythagorean polygons for which the perimeter is  n. +For a given integer n, define P(n) as the number of distinct pythagorean polygons for which the perimeter is  n. Pythagorean polygons should be considered distinct as long as none is a translation of another. -You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045. +You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045. Find P(120). diff --git a/project_euler/problems/293_pseudofortunate_numbers.txt b/project_euler/problems/293_pseudofortunate_numbers.txt index d36daab..d401abf 100644 --- a/project_euler/problems/293_pseudofortunate_numbers.txt +++ b/project_euler/problems/293_pseudofortunate_numbers.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=293 Pseudo-Fortunate Numbers - -An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes. -The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48. - -If N is admissible, the smallest integer M 1 such that N+M is prime, will be called the pseudo-Fortunate number for N. +An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes. +The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48. - -For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7. -The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11. -It can also be seen that the pseudo-Fortunate number for 16 is 3. - -Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 109. +If N is admissible, the smallest integer M 1 such that N+M is prime, will be called the pseudo-Fortunate number for N. + + +For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7. +The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11. +It can also be seen that the pseudo-Fortunate number for 16 is 3. + + +Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 109. diff --git a/project_euler/problems/294_sum_of_digits_experience_23.txt b/project_euler/problems/294_sum_of_digits_experience_23.txt index 5e2dc9c..dd9c182 100644 --- a/project_euler/problems/294_sum_of_digits_experience_23.txt +++ b/project_euler/problems/294_sum_of_digits_experience_23.txt @@ -2,20 +2,20 @@ http://projecteuler.net/problem=294 Sum of digits - experience #23 - -For a positive integer k, define d(k) as the sum of the digits of k in its usual decimal representation. -Thus d(42) = 4+2 = 6. - - -For a positive integer n, define S(n) as the number of positive integers k n with the following properties : - -k is divisible by 23 and -d(k) = 23. - - -You are given that S(9) = 263626 and S(42) = 6377168878570056. - - -Find S(1112) and give your answer mod 109. + +For a positive integer k, define d(k) as the sum of the digits of k in its usual decimal representation. +Thus d(42) = 4+2 = 6. + + +For a positive integer n, define S(n) as the number of positive integers k n with the following properties : + +k is divisible by 23 and +d(k) = 23. + + +You are given that S(9) = 263626 and S(42) = 6377168878570056. + + +Find S(1112) and give your answer mod 109. diff --git a/project_euler/problems/295_lenticular_holes.txt b/project_euler/problems/295_lenticular_holes.txt index 54c0ade..5c4c49f 100644 --- a/project_euler/problems/295_lenticular_holes.txt +++ b/project_euler/problems/295_lenticular_holes.txt @@ -2,30 +2,30 @@ http://projecteuler.net/problem=295 Lenticular holes -We call the convex area enclosed by two circles a lenticular hole if: +We call the convex area enclosed by two circles a lenticular hole if: The centres of both circles are on lattice points. The two circles intersect at two distinct lattice points. -The interior of the convex area enclosed by both circles does not contain any lattice points. +The interior of the convex area enclosed by both circles does not contain any lattice points. -Consider the circles: -C0: x2+y2=25 -C1: (x+4)2+(y-4)2=1 -C2: (x-12)2+(y-4)2=65 +Consider the circles: +C0: x2+y2=25 +C1: (x+4)2+(y-4)2=1 +C2: (x-12)2+(y-4)2=65 + - The circles C0, C1 and C2 are drawn in the picture below. - + C0 and C1 form a lenticular hole, as well as C0 and C2. - -We call an ordered pair of positive real numbers (r1, r2) a lenticular pair if there exist two circles with radii r1 and r2 that form a lenticular hole. + +We call an ordered pair of positive real numbers (r1, r2) a lenticular pair if there exist two circles with radii r1 and r2 that form a lenticular hole. We can verify that (1, 5) and (5, 65) are the lenticular pairs of the example above. - -Let L(N) be the number of distinct lenticular pairs (r1, r2) for which 0 r1 r2 N. + +Let L(N) be the number of distinct lenticular pairs (r1, r2) for which 0 r1 r2 N. We can verify that L(10) = 30 and L(100) = 3442. - -Find L(100 000). + +Find L(100 000). diff --git a/project_euler/problems/296_angular_bisector_and_tangent.txt b/project_euler/problems/296_angular_bisector_and_tangent.txt index ad7415d..2ea8911 100644 --- a/project_euler/problems/296_angular_bisector_and_tangent.txt +++ b/project_euler/problems/296_angular_bisector_and_tangent.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=296 Angular Bisector and Tangent - + Given is an integer sided triangle ABC with BC AC AB. -k is the angular bisector of angle ACB. m is the tangent at C to the circumscribed circle of ABC. n is a line parallel to m through B. -The intersection of n and k is called E. +k is the angular bisector of angle ACB. m is the tangent at C to the circumscribed circle of ABC. n is a line parallel to m through B. +The intersection of n and k is called E. + - -How many triangles ABC with a perimeter not exceeding 100 000 exist such that BE has integral length? +How many triangles ABC with a perimeter not exceeding 100 000 exist such that BE has integral length? diff --git a/project_euler/problems/297_zeckendorf_representation.txt b/project_euler/problems/297_zeckendorf_representation.txt index c42791f..0136cdf 100644 --- a/project_euler/problems/297_zeckendorf_representation.txt +++ b/project_euler/problems/297_zeckendorf_representation.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=297 Zeckendorf Representation -Each new term in the Fibonacci sequence is generated by adding the previous two terms. +Each new term in the Fibonacci sequence is generated by adding the previous two terms. Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89. -Every positive integer can be uniquely written as a sum of nonconsecutive terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89. +Every positive integer can be uniquely written as a sum of nonconsecutive terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89. Such a sum is called the Zeckendorf representation of the number. -For any integer n>0, let z(n) be the number of terms in the Zeckendorf representation of n. -Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc. +For any integer n>0, let z(n) be the number of terms in the Zeckendorf representation of n. +Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc. Also, for 0n106, ∑ z(n) = 7894453. Find ∑ z(n) for 0n1017. diff --git a/project_euler/problems/298_selective_amnesia.txt b/project_euler/problems/298_selective_amnesia.txt index e425ad2..da4df88 100644 --- a/project_euler/problems/298_selective_amnesia.txt +++ b/project_euler/problems/298_selective_amnesia.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=298 Selective Amnesia - -table.p298, table.p298 th, table.p298 td { - border-width: 1px 1px 1px 1px; - border-style: solid solid solid solid; - border-color: black black black black; - text-align:center; - -moz-border-radius: 0px 0px 0px 0px; -} -table.p298 { - border-spacing: 1px; - border-collapse: separate; - background-color: rgb(255,255,255); -} -table.p298 th, table.p298 td { - padding: 1px 6px 1px 6px; -} -table.p298 th { background-color: rgb(200,220,250); } -table.p298 td { background-color: rgb(255,255,255); } + +table.p298, table.p298 th, table.p298 td { + border-width: 1px 1px 1px 1px; + border-style: solid solid solid solid; + border-color: black black black black; + text-align:center; + -moz-border-radius: 0px 0px 0px 0px; +} +table.p298 { + border-spacing: 1px; + border-collapse: separate; + background-color: rgb(255,255,255); +} +table.p298 th, table.p298 td { + padding: 1px 6px 1px 6px; +} +table.p298 th { background-color: rgb(200,220,250); } +table.p298 td { background-color: rgb(255,255,255); } Larry and Robin play a memory game involving of a sequence of random numbers between 1 and 10, inclusive, that are called out one at a time. Each player can remember up to 5 previous numbers. When the called number is in a player's memory, that player is awarded a point. If it's not, the player adds the called number to his memory, removing another number if his memory is full. -Both players start with empty memories. Both players always add new missed numbers to their memory but use a different strategy in deciding which number to remove: -Larry's strategy is to remove the number that hasn't been called in the longest time. +Both players start with empty memories. Both players always add new missed numbers to their memory but use a different strategy in deciding which number to remove: +Larry's strategy is to remove the number that hasn't been called in the longest time. Robin's strategy is to remove the number that's been in the memory the longest time. Example game: diff --git a/project_euler/problems/299_three_similar_triangles.txt b/project_euler/problems/299_three_similar_triangles.txt index 2df886a..5538c9f 100644 --- a/project_euler/problems/299_three_similar_triangles.txt +++ b/project_euler/problems/299_three_similar_triangles.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=299 Three similar triangles -Four points with integer coordinates are selected:A(a, 0), B(b, 0), C(0, c) and D(0, d), -with 0  a  b and 0  c  d. +Four points with integer coordinates are selected:A(a, 0), B(b, 0), C(0, c) and D(0, d), +with 0  a  b and 0  c  d. Point P, also with integer coordinates, is chosen on the line AC so that the three triangles ABP, CDP and BDP are all similar. It is easy to prove that the three triangles can be similar, only if a=c. So, given that a=c, we are looking for triplets (a,b,d) such that at least one point P (with integer coordinates) exists on AC, making the three triangles ABP, CDP and BDP all similar. -For example, if (a,b,d)=(2,3,4), it can be easily verified that point P(1,1) satisfies the above condition. +For example, if (a,b,d)=(2,3,4), it can be easily verified that point P(1,1) satisfies the above condition. Note that the triplets (2,3,4) and (2,4,3) are considered as distinct, although point P(1,1) is common for both. -If b+d  100, there are 92 distinct triplets (a,b,d) such that point P exists. +If b+d  100, there are 92 distinct triplets (a,b,d) such that point P exists. If b+d  100 000, there are 320471 distinct triplets (a,b,d) such that point P exists. If b+d  100 000 000, how many distinct triplets (a,b,d) are there such that point P exists? diff --git a/project_euler/problems/300_protein_folding.txt b/project_euler/problems/300_protein_folding.txt index accb927..629cbff 100644 --- a/project_euler/problems/300_protein_folding.txt +++ b/project_euler/problems/300_protein_folding.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=300 Protein folding -In a very simplified form, we can consider proteins as strings consisting of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH. +In a very simplified form, we can consider proteins as strings consisting of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH. For this problem, the orientation of a protein is important; e.g. HPP is considered distinct from PPH. Thus, there are 2n distinct proteins consisting of n elements. -When one encounters these strings in nature, they are always folded in such a way that the number of H-H contact points is as large as possible, since this is energetically advantageous. -As a result, the H-elements tend to accumulate in the inner part, with the P-elements on the outside. +When one encounters these strings in nature, they are always folded in such a way that the number of H-H contact points is as large as possible, since this is energetically advantageous. +As a result, the H-elements tend to accumulate in the inner part, with the P-elements on the outside. Natural proteins are folded in three dimensions of course, but we will only consider protein folding in two dimensions. The figure below shows two possible ways that our example protein could be folded (H-H contact points are shown with red dots). -The folding on the left has only six H-H contact points, thus it would never occur naturally. +The folding on the left has only six H-H contact points, thus it would never occur naturally. On the other hand, the folding on the right has nine H-H contact points, which is optimal for this string. Assuming that H and P elements are equally likely to occur in any position along the string, the average number of H-H contact points in an optimal folding of a random protein string of length 8 turns out to be 850 / 28=3.3203125. -What is the average number of H-H contact points in an optimal folding of a random protein string of length 15? +What is the average number of H-H contact points in an optimal folding of a random protein string of length 15? Give your answer using as many decimal places as necessary for an exact result. diff --git a/project_euler/problems/301_nim.txt b/project_euler/problems/301_nim.txt index 8e0206a..3b8bc45 100644 --- a/project_euler/problems/301_nim.txt +++ b/project_euler/problems/301_nim.txt @@ -3,18 +3,18 @@ http://projecteuler.net/problem=301 Nim Nim is a game played with heaps of stones, where two players take it in turn to remove any number of stones from any heap until no stones remain. -We'll consider the three-heap normal-play version of Nim, which works as follows: -- At the start of the game there are three heaps of stones. -- On his turn the player removes any positive number of stones from any single heap. +We'll consider the three-heap normal-play version of Nim, which works as follows: +- At the start of the game there are three heaps of stones. +- On his turn the player removes any positive number of stones from any single heap. - The first player unable to move (because no stones remain) loses. - If (n1,n2,n3) indicates a Nim position consisting of heaps of size n1, n2 and n3 then there is a simple function X(n1,n2,n3) — that you may look up or attempt to deduce for yourself — that returns: + If (n1,n2,n3) indicates a Nim position consisting of heaps of size n1, n2 and n3 then there is a simple function X(n1,n2,n3) — that you may look up or attempt to deduce for yourself — that returns: zero if, with perfect strategy, the player about to move will eventually lose; or non-zero if, with perfect strategy, the player about to move will eventually win. -For example X(1,2,3) = 0 because, no matter what the current player does, his opponent can respond with a move that leaves two heaps of equal size, at which point every move by the current player can be mirrored by his opponent until no stones remain; so the current player loses. To illustrate: -- current player moves to (1,2,1) -- opponent moves to (1,0,1) -- current player moves to (0,0,1) +For example X(1,2,3) = 0 because, no matter what the current player does, his opponent can respond with a move that leaves two heaps of equal size, at which point every move by the current player can be mirrored by his opponent until no stones remain; so the current player loses. To illustrate: +- current player moves to (1,2,1) +- opponent moves to (1,0,1) +- current player moves to (0,0,1) - opponent moves to (0,0,0), and so wins. -For how many positive integers n  230 does X(n,2n,3n) = 0 ? +For how many positive integers n  230 does X(n,2n,3n) = 0 ? diff --git a/project_euler/problems/302_strong_achilles_numbers.txt b/project_euler/problems/302_strong_achilles_numbers.txt index 6069cc1..cface00 100644 --- a/project_euler/problems/302_strong_achilles_numbers.txt +++ b/project_euler/problems/302_strong_achilles_numbers.txt @@ -2,25 +2,25 @@ http://projecteuler.net/problem=302 Strong Achilles Numbers - -A positive integer n is powerful if p2 is a divisor of n for every prime factor p in n. - -A positive integer n is a perfect power if n can be expressed as a power of another positive integer. +A positive integer n is powerful if p2 is a divisor of n for every prime factor p in n. - -A positive integer n is an Achilles number if n is powerful but not a perfect power. For example, 864 and 1800 are Achilles numbers: 864 = 25·33 and 1800 = 23·32·52. - -We shall call a positive integer S a Strong Achilles number if both S and φ(S) are Achilles numbers.1 -For example, 864 is a Strong Achilles number: φ(864) = 288 = 25·32. However, 1800 isn't a Strong Achilles number because: φ(1800) = 480 = 25·31·51. +A positive integer n is a perfect power if n can be expressed as a power of another positive integer. -There are 7 Strong Achilles numbers below 104 and 656 below 108. - -How many Strong Achilles numbers are there below 1018? +A positive integer n is an Achilles number if n is powerful but not a perfect power. For example, 864 and 1800 are Achilles numbers: 864 = 25·33 and 1800 = 23·32·52. -1 φ denotes Euler's totient function. +We shall call a positive integer S a Strong Achilles number if both S and φ(S) are Achilles numbers.1 +For example, 864 is a Strong Achilles number: φ(864) = 288 = 25·32. However, 1800 isn't a Strong Achilles number because: φ(1800) = 480 = 25·31·51. + +There are 7 Strong Achilles numbers below 104 and 656 below 108. + + +How many Strong Achilles numbers are there below 1018? + + +1 φ denotes Euler's totient function. diff --git a/project_euler/problems/303_multiples_with_small_digits.txt b/project_euler/problems/303_multiples_with_small_digits.txt index e542790..ffb740b 100644 --- a/project_euler/problems/303_multiples_with_small_digits.txt +++ b/project_euler/problems/303_multiples_with_small_digits.txt @@ -2,11 +2,11 @@ http://projecteuler.net/problem=303 Multiples with small digits - + For a positive integer n, define f(n) as the least positive multiple of n that, written in base 10, uses only digits 2. Thus f(2)=2, f(3)=12, f(7)=21, f(42)=210, f(89)=1121222. Also, . - -Find . + +Find . diff --git a/project_euler/problems/304_primonacci.txt b/project_euler/problems/304_primonacci.txt index 119a4f7..383c71b 100644 --- a/project_euler/problems/304_primonacci.txt +++ b/project_euler/problems/304_primonacci.txt @@ -2,26 +2,26 @@ http://projecteuler.net/problem=304 Primonacci - -For any positive integer n the function next_prime(n) returns the smallest prime p such that pn. - - -The sequence a(n) is defined by: -a(1)=next_prime(1014) and a(n)=next_prime(a(n-1)) for n1. - - -The fibonacci sequence f(n) is defined by: -f(0)=0, f(1)=1 and f(n)=f(n-1)+f(n-2) for n1. - - -The sequence b(n) is defined as f(a(n)). - - -Find b(n) for 1n100 000. -Give your answer mod 1234567891011. - - - - - + +For any positive integer n the function next_prime(n) returns the smallest prime p such that pn. + + +The sequence a(n) is defined by: +a(1)=next_prime(1014) and a(n)=next_prime(a(n-1)) for n1. + + +The fibonacci sequence f(n) is defined by: +f(0)=0, f(1)=1 and f(n)=f(n-1)+f(n-2) for n1. + + +The sequence b(n) is defined as f(a(n)). + + +Find b(n) for 1n100 000. +Give your answer mod 1234567891011. + + + + + diff --git a/project_euler/problems/305_reflexive_position.txt b/project_euler/problems/305_reflexive_position.txt index 2ce824b..f1fb3c8 100644 --- a/project_euler/problems/305_reflexive_position.txt +++ b/project_euler/problems/305_reflexive_position.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=305 Reflexive Position - -Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10. -Thus, S = 1234567891011121314151617181920212223242... - -It's easy to see that any number will show up an infinite number of times in S. +Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10. +Thus, S = 1234567891011121314151617181920212223242... - -Let's call f(n) the starting position of the nth occurrence of n in S. -For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365. - -Find f(3k) for 1k13. +It's easy to see that any number will show up an infinite number of times in S. + + +Let's call f(n) the starting position of the nth occurrence of n in S. +For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365. + + +Find f(3k) for 1k13. diff --git a/project_euler/problems/306_paperstrip_game.txt b/project_euler/problems/306_paperstrip_game.txt index 6b5e5c0..0845c5b 100644 --- a/project_euler/problems/306_paperstrip_game.txt +++ b/project_euler/problems/306_paperstrip_game.txt @@ -3,8 +3,8 @@ http://projecteuler.net/problem=306 Paper-strip Game The following game is a classic example of Combinatorial Game Theory: -Two players start with a strip of n white squares and they take alternate turns. -On each turn, a player picks two contiguous white squares and paints them black. +Two players start with a strip of n white squares and they take alternate turns. +On each turn, a player picks two contiguous white squares and paints them black. The first player who cannot make a move loses. If n = 1, there are no valid moves, so the first player loses automatically. @@ -14,7 +14,7 @@ If n = 4, there are three valid moves for the first player; she can win the game If n = 5, there are four valid moves for the first player (shown below in red); but no matter what she does, the second player (blue) wins. -So, for 1 n 5, there are 3 values of n for which the first player can force a win. +So, for 1 n 5, there are 3 values of n for which the first player can force a win. Similarly, for 1 n 50, there are 40 values of n for which the first player can force a win. For 1 n 1 000 000, how many values of n are there for which the first player can force a win? diff --git a/project_euler/problems/307_chip_defects.txt b/project_euler/problems/307_chip_defects.txt index 9ef71cc..1226815 100644 --- a/project_euler/problems/307_chip_defects.txt +++ b/project_euler/problems/307_chip_defects.txt @@ -3,13 +3,13 @@ http://projecteuler.net/problem=307 Chip Defects -k defects are randomly distributed amongst n integrated-circuit chips produced by a factory (any number of defects may be found on a chip and each defect is independent of the other defects). +k defects are randomly distributed amongst n integrated-circuit chips produced by a factory (any number of defects may be found on a chip and each defect is independent of the other defects). - -Let p(k,n) represent the probability that there is a chip with at least 3 defects. -For instance p(3,7) 0.0204081633. - -Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal places in the form 0.abcdefghij +Let p(k,n) represent the probability that there is a chip with at least 3 defects. +For instance p(3,7) 0.0204081633. + + +Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal places in the form 0.abcdefghij diff --git a/project_euler/problems/308_an_amazing_primegenerating_automaton.txt b/project_euler/problems/308_an_amazing_primegenerating_automaton.txt index dadc968..195fbba 100644 --- a/project_euler/problems/308_an_amazing_primegenerating_automaton.txt +++ b/project_euler/problems/308_an_amazing_primegenerating_automaton.txt @@ -65,9 +65,9 @@ For example, one of the Fractran programs that John Horton Conway wrote for prim . -Starting with the seed integer 2, successive iterations of the program produce the sequence: +Starting with the seed integer 2, successive iterations of the program produce the sequence: 15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544, 32, 240, ... -The powers of 2 that appear in this sequence are 22, 23, 25, ... +The powers of 2 that appear in this sequence are 22, 23, 25, ... It can be shown that all the powers of 2 in this sequence have prime exponents and that all the primes appear as exponents of powers of 2, in proper order! -If someone uses the above Fractran program to solve Project Euler Problem 7 (find the 10001st prime), how many iterations would be needed until the program produces 210001st prime ? +If someone uses the above Fractran program to solve Project Euler Problem 7 (find the 10001st prime), how many iterations would be needed until the program produces 210001st prime ? diff --git a/project_euler/problems/309_integer_ladders.txt b/project_euler/problems/309_integer_ladders.txt index 8eeb853..52e3077 100644 --- a/project_euler/problems/309_integer_ladders.txt +++ b/project_euler/problems/309_integer_ladders.txt @@ -4,9 +4,9 @@ Integer Ladders In the classic "Crossing Ladders" problem, we are given the lengths x and y of two ladders resting on the opposite walls of a narrow, level street. We are also given the height h above the street where the two ladders cross and we are asked to find the width of the street (w). -Here, we are only concerned with instances where all four variables are positive integers. +Here, we are only concerned with instances where all four variables are positive integers. For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56. -In fact, for integer values x, y, h and 0 x y x,y,h) producing integer solutions for w: +In fact, for integer values x, y, h and 0 x y x,y,h) producing integer solutions for w: (70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175, 40). For integer values x, y, h and 0 x y x,y,h) produce integer solutions for w? diff --git a/project_euler/problems/310_nim_square.txt b/project_euler/problems/310_nim_square.txt index fbd589c..e948961 100644 --- a/project_euler/problems/310_nim_square.txt +++ b/project_euler/problems/310_nim_square.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=310 Nim Square - -Alice and Bob play the game Nim Square. -Nim Square is just like ordinary three-heap normal play Nim, but the players may only remove a square number of stones from a heap. -The number of stones in the three heaps is represented by the ordered triple (a,b,c). -If 0abc29 then the number of losing positions for the next player is 1160. - - -Find the number of losing positions for the next player if 0abc100 000. + +Alice and Bob play the game Nim Square. +Nim Square is just like ordinary three-heap normal play Nim, but the players may only remove a square number of stones from a heap. +The number of stones in the three heaps is represented by the ordered triple (a,b,c). +If 0abc29 then the number of losing positions for the next player is 1160. + + +Find the number of losing positions for the next player if 0abc100 000. diff --git a/project_euler/problems/311_biclinic_integral_quadrilaterals.txt b/project_euler/problems/311_biclinic_integral_quadrilaterals.txt index fe20776..3fd23d0 100644 --- a/project_euler/problems/311_biclinic_integral_quadrilaterals.txt +++ b/project_euler/problems/311_biclinic_integral_quadrilaterals.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=311 Biclinic Integral Quadrilaterals -ABCD is a convex, integer sided quadrilateral with 1 AB BC CD AD. -BD has integer length. O is the midpoint of BD. AO has integer length. +ABCD is a convex, integer sided quadrilateral with 1 AB BC CD AD. +BD has integer length. O is the midpoint of BD. AO has integer length. We'll call ABCD a biclinic integral quadrilateral if AO = CO BO = DO. -For example, the following quadrilateral is a biclinic integral quadrilateral: -AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23. +For example, the following quadrilateral is a biclinic integral quadrilateral: +AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23. -Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD that satisfy AB2+BC2+CD2+AD2 N. -We can verify that B(10 000) = 49 and B(1 000 000) = 38239. +Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD that satisfy AB2+BC2+CD2+AD2 N. +We can verify that B(10 000) = 49 and B(1 000 000) = 38239. -Find B(10 000 000 000). +Find B(10 000 000 000). diff --git a/project_euler/problems/312_cyclic_paths_on_sierpiski_graphs.txt b/project_euler/problems/312_cyclic_paths_on_sierpiski_graphs.txt index 44616e5..b38d56a 100644 --- a/project_euler/problems/312_cyclic_paths_on_sierpiski_graphs.txt +++ b/project_euler/problems/312_cyclic_paths_on_sierpiski_graphs.txt @@ -1,22 +1,22 @@ http://projecteuler.net/problem=312 -Cyclic paths on Sierpiński graphs +Cyclic paths on Sierpiński graphs -- A Sierpiński graph of order-1 (S1) is an equilateral triangle. -- Sn+1 is obtained from Sn by positioning three copies of Sn so that every pair of copies has one common corner. +- A Sierpiński graph of order-1 (S1) is an equilateral triangle. +- Sn+1 is obtained from Sn by positioning three copies of Sn so that every pair of copies has one common corner. -Let C(n) be the number of cycles that pass exactly once through all the vertices of Sn. -For example, C(3) = 8 because eight such cycles can be drawn on S3, as shown below: +Let C(n) be the number of cycles that pass exactly once through all the vertices of Sn. +For example, C(3) = 8 because eight such cycles can be drawn on S3, as shown below: -It can also be verified that : -C(1) = C(2) = 1 -C(5) = 71328803586048 -C(10 000) mod 108 = 37652224 +It can also be verified that : +C(1) = C(2) = 1 +C(5) = 71328803586048 +C(10 000) mod 108 = 37652224 C(10 000) mod 138 = 617720485 -Find C(C(C(10 000))) mod 138. +Find C(C(C(10 000))) mod 138. diff --git a/project_euler/problems/314_the_mouse_on_the_moon.txt b/project_euler/problems/314_the_mouse_on_the_moon.txt index d3f4c24..1971884 100644 --- a/project_euler/problems/314_the_mouse_on_the_moon.txt +++ b/project_euler/problems/314_the_mouse_on_the_moon.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=314 The Mouse on the Moon - -The moon has been opened up, and land can be obtained for free, but there is a catch. You have to build a wall around the land that you stake out, and building a wall on the moon is expensive. Every country has been allotted a 500 m by 500 m square area, but they will possess only that area which they wall in. 251001 posts have been placed in a rectangular grid with 1 meter spacing. The wall must be a closed series of straight lines, each line running from post to post. - -The bigger countries of course have built a 2000 m wall enclosing the entire 250 000 m2 area. The Duchy of Grand Fenwick, has a tighter budget, and has asked you (their Royal Programmer) to compute what shape would get best maximum enclosed-area/wall-length ratio. +The moon has been opened up, and land can be obtained for free, but there is a catch. You have to build a wall around the land that you stake out, and building a wall on the moon is expensive. Every country has been allotted a 500 m by 500 m square area, but they will possess only that area which they wall in. 251001 posts have been placed in a rectangular grid with 1 meter spacing. The wall must be a closed series of straight lines, each line running from post to post. - -You have done some preliminary calculations on a sheet of paper. -For a 2000 meter wall enclosing the 250 000 m2 area the -enclosed-area/wall-length ratio is 125. -Although not allowed , but to get an idea if this is anything better: if you place a circle inside the square area touching the four sides the area will be equal to π*2502 m2 and the perimeter will be π*500 m, so the enclosed-area/wall-length ratio will also be 125. - -However, if you cut off from the square four triangles with sides 75 m, 75 m and 752 m the total area becomes 238750 m2 and the perimeter becomes 1400+3002 m. So this gives an enclosed-area/wall-length ratio of 130.87, which is significantly better. +The bigger countries of course have built a 2000 m wall enclosing the entire 250 000 m2 area. The Duchy of Grand Fenwick, has a tighter budget, and has asked you (their Royal Programmer) to compute what shape would get best maximum enclosed-area/wall-length ratio. - -Find the maximum enclosed-area/wall-length ratio. -Give your answer rounded to 8 places behind the decimal point in the form abc.defghijk. +You have done some preliminary calculations on a sheet of paper. +For a 2000 meter wall enclosing the 250 000 m2 area the +enclosed-area/wall-length ratio is 125. +Although not allowed , but to get an idea if this is anything better: if you place a circle inside the square area touching the four sides the area will be equal to π*2502 m2 and the perimeter will be π*500 m, so the enclosed-area/wall-length ratio will also be 125. + + +However, if you cut off from the square four triangles with sides 75 m, 75 m and 752 m the total area becomes 238750 m2 and the perimeter becomes 1400+3002 m. So this gives an enclosed-area/wall-length ratio of 130.87, which is significantly better. + + + +Find the maximum enclosed-area/wall-length ratio. +Give your answer rounded to 8 places behind the decimal point in the form abc.defghijk. diff --git a/project_euler/problems/315_digital_root_clocks.txt b/project_euler/problems/315_digital_root_clocks.txt index 1d0d614..09c7791 100644 --- a/project_euler/problems/315_digital_root_clocks.txt +++ b/project_euler/problems/315_digital_root_clocks.txt @@ -3,16 +3,16 @@ http://projecteuler.net/problem=315 Digital root clocks -Sam and Max are asked to transform two digital clocks into two "digital root" clocks. +Sam and Max are asked to transform two digital clocks into two "digital root" clocks. A digital root clock is a digital clock that calculates digital roots step by step. -When a clock is fed a number, it will show it and then it will start the calculation, showing all the intermediate values until it gets to the result. +When a clock is fed a number, it will show it and then it will start the calculation, showing all the intermediate values until it gets to the result. For example, if the clock is fed the number 137, it will show: "137" "11" "2" and then it will go black, waiting for the next number. -Every digital number consists of some light segments: three horizontal (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, bottom-right). +Every digital number consists of some light segments: three horizontal (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, bottom-right). Number "1" is made of vertical top-right and bottom-right, number "4" is made by middle horizontal and vertical top-left, top-right and bottom-right. Number "8" lights them all. -The clocks consume energy only when segments are turned on/off. +The clocks consume energy only when segments are turned on/off. To turn on a "2" will cost 5 transitions, while a "7" will cost only 4 transitions. Sam and Max built two different clocks. -Sam's clock is fed e.g. number 137: the clock shows "137", then the panel is turned off, then the next number ("11") is turned on, then the panel is turned off again and finally the last number ("2") is turned on and, after some time, off. +Sam's clock is fed e.g. number 137: the clock shows "137", then the panel is turned off, then the next number ("11") is turned on, then the panel is turned off again and finally the last number ("2") is turned on and, after some time, off. For the example, with number 137, Sam's clock requires: "137" @@ -26,31 +26,31 @@ For the example, with number 137, Sam's clock requires: "2" : (5) 2 = 10 transitions ("2" on/off). - -For a grand total of 40 transitions. - -Max's clock works differently. Instead of turning off the whole panel, it is smart enough to turn off only those segments that won't be needed for the next number. + +For a grand total of 40 transitions. + +Max's clock works differently. Instead of turning off the whole panel, it is smart enough to turn off only those segments that won't be needed for the next number. For number 137, Max's clock requires: "137" : -2 + 5 + 4 = 11 transitions ("137" on) +2 + 5 + 4 = 11 transitions ("137" on) 7 transitions (to turn off the segments that are not needed for number "11"). "11" : -0 transitions (number "11" is already turned on correctly) -3 transitions (to turn off the first "1" and the bottom part of the second "1"; +0 transitions (number "11" is already turned on correctly) +3 transitions (to turn off the first "1" and the bottom part of the second "1"; the top part is common with number "2"). "2" : -4 tansitions (to turn on the remaining segments in order to get a "2") +4 tansitions (to turn on the remaining segments in order to get a "2") 5 transitions (to turn off number "2"). - -For a grand total of 30 transitions. - -Of course, Max's clock consumes less power than Sam's one. -The two clocks are fed all the prime numbers between A = 107 and B = 2107. + +For a grand total of 30 transitions. + +Of course, Max's clock consumes less power than Sam's one. +The two clocks are fed all the prime numbers between A = 107 and B = 2107. Find the difference between the total number of transitions needed by Sam's clock and that needed by Max's one. diff --git a/project_euler/problems/316_numbers_in_decimal_expansions.txt b/project_euler/problems/316_numbers_in_decimal_expansions.txt index 041dc81..a6f9c3e 100644 --- a/project_euler/problems/316_numbers_in_decimal_expansions.txt +++ b/project_euler/problems/316_numbers_in_decimal_expansions.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=316 Numbers in decimal expansions -Let p = p1 p2 p3 ... be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. -It can be seen that p corresponds to the real number 0.p1 p2 p3 .... +Let p = p1 p2 p3 ... be an infinite sequence of random digits, selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. +It can be seen that p corresponds to the real number 0.p1 p2 p3 .... It can also be seen that choosing a random real number from the interval [0,1) is equivalent to choosing an infinite sequence of random digits selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. -For any positive integer n with d decimal digits, let k be the smallest index such that pk, pk+1, ...pk+d-1 are the decimal digits of n, in the same order. +For any positive integer n with d decimal digits, let k be the smallest index such that pk, pk+1, ...pk+d-1 are the decimal digits of n, in the same order. Also, let g(n) be the expected value of k; it can be proven that g(n) is always finite and, interestingly, always an integer number. -For example, if n = 535, then -for p = 31415926535897...., we get k = 9 -for p = 355287143650049560000490848764084685354..., we get k = 36 +For example, if n = 535, then +for p = 31415926535897...., we get k = 9 +for p = 355287143650049560000490848764084685354..., we get k = 36 etc and we find that g(535) = 1008. -Given that , find -Note: represents the floor function. +Given that , find +Note: represents the floor function. diff --git a/project_euler/problems/317_firecracker.txt b/project_euler/problems/317_firecracker.txt index 7a2be7b..7c3be11 100644 --- a/project_euler/problems/317_firecracker.txt +++ b/project_euler/problems/317_firecracker.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=317 Firecracker - -A firecracker explodes at a height of 100 m above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of 20 m/s. - -We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s2. +A firecracker explodes at a height of 100 m above level ground. It breaks into a large number of very small fragments, which move in every direction; all of them have the same initial velocity of 20 m/s. - -Find the volume (in m3) of the region through which the fragments move before reaching the ground. -Give your answer rounded to four decimal places. + +We assume that the fragments move without air resistance, in a uniform gravitational field with g=9.81 m/s2. + + +Find the volume (in m3) of the region through which the fragments move before reaching the ground. +Give your answer rounded to four decimal places. diff --git a/project_euler/problems/318_2011_nines.txt b/project_euler/problems/318_2011_nines.txt index efd06e7..ffdb4f1 100644 --- a/project_euler/problems/318_2011_nines.txt +++ b/project_euler/problems/318_2011_nines.txt @@ -2,34 +2,34 @@ http://projecteuler.net/problem=318 2011 nines - -Consider the real number 2+3. -When we calculate the even powers of 2+3 -we get: -(2+3)2 = 9.898979485566356... -(2+3)4 = 97.98979485566356... -(2+3)6 = 969.998969071069263... -(2+3)8 = 9601.99989585502907... -(2+3)10 = 95049.999989479221... -(2+3)12 = 940897.9999989371855... -(2+3)14 = 9313929.99999989263... + +Consider the real number 2+3. +When we calculate the even powers of 2+3 +we get: +(2+3)2 = 9.898979485566356... +(2+3)4 = 97.98979485566356... +(2+3)6 = 969.998969071069263... +(2+3)8 = 9601.99989585502907... +(2+3)10 = 95049.999989479221... +(2+3)12 = 940897.9999989371855... +(2+3)14 = 9313929.99999989263... (2+3)16 = 92198401.99999998915... - -It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. -In fact it can be proven that the fractional part of (2+3)2n approaches 1 for large n. - -Consider all real numbers of the form p+q with p and q positive integers and pq, such that the fractional part -of (p+q)2n approaches 1 for large n. +It looks like that the number of consecutive nines at the beginning of the fractional part of these powers is non-decreasing. +In fact it can be proven that the fractional part of (2+3)2n approaches 1 for large n. + + +Consider all real numbers of the form p+q with p and q positive integers and pq, such that the fractional part +of (p+q)2n approaches 1 for large n. + + +Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (p+q)2n. + - -Let C(p,q,n) be the number of consecutive nines at the beginning of the fractional part of (p+q)2n. +Let N(p,q) be the minimal value of n such that C(p,q,n) 2011. - -Let N(p,q) be the minimal value of n such that C(p,q,n) 2011. - -Find N(p,q) for p+q 2011. +Find N(p,q) for p+q 2011. diff --git a/project_euler/problems/319_bounded_sequences.txt b/project_euler/problems/319_bounded_sequences.txt index 83da4ac..67c1ada 100644 --- a/project_euler/problems/319_bounded_sequences.txt +++ b/project_euler/problems/319_bounded_sequences.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=319 Bounded Sequences - -Let x1, x2,..., xn be a sequence of length n such that: + +Let x1, x2,..., xn be a sequence of length n such that: x1 = 2 for all 1 i n : xi-1 xi for all i and j with 1 i, j n : (xi) j (xj + 1)i - -There are only five such sequences of length 2, namely: -{2,4}, {2,5}, {2,6}, {2,7} and {2,8}. -There are 293 such sequences of length 5; three examples are given below: -{2,5,11,25,55}, {2,6,14,36,88}, {2,8,22,64,181}. - -Let t(n) denote the number of such sequences of length n. -You are given that t(10) = 86195 and t(20) = 5227991891. +There are only five such sequences of length 2, namely: +{2,4}, {2,5}, {2,6}, {2,7} and {2,8}. +There are 293 such sequences of length 5; three examples are given below: +{2,5,11,25,55}, {2,6,14,36,88}, {2,8,22,64,181}. + + +Let t(n) denote the number of such sequences of length n. +You are given that t(10) = 86195 and t(20) = 5227991891. + - -Find t(1010) and give your answer modulo 109. +Find t(1010) and give your answer modulo 109. diff --git a/project_euler/problems/320_factorials_divisible_by_a_huge_integer.txt b/project_euler/problems/320_factorials_divisible_by_a_huge_integer.txt index 703ad35..20fca0f 100644 --- a/project_euler/problems/320_factorials_divisible_by_a_huge_integer.txt +++ b/project_euler/problems/320_factorials_divisible_by_a_huge_integer.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=320 Factorials divisible by a huge integer - + Let N(i) be the smallest integer n such that n! is divisible by (i!)1234567890 - -Let S(u)=N(i) for 10 i u. - -S(1000)=614538266565663. +Let S(u)=N(i) for 10 i u. + + +S(1000)=614538266565663. + - -Find S(1 000 000) mod 1018. +Find S(1 000 000) mod 1018. diff --git a/project_euler/problems/321_swapping_counters.txt b/project_euler/problems/321_swapping_counters.txt index 6e30cc1..08d4132 100644 --- a/project_euler/problems/321_swapping_counters.txt +++ b/project_euler/problems/321_swapping_counters.txt @@ -8,7 +8,7 @@ A counter can move from one square to the next (slide) or can jump over another Let M(n) represent the minimum number of moves/actions to completely reverse the positions of the coloured counters; that is, move all the red counters to the right and all the blue counters to the left. It can be verified M(3) = 15, which also happens to be a triangle number. -If we create a sequence based on the values of n for which M(n) is a triangle number then the first five terms would be: +If we create a sequence based on the values of n for which M(n) is a triangle number then the first five terms would be: 1, 3, 10, 22, and 63, and their sum would be 99. Find the sum of the first forty terms of this sequence. diff --git a/project_euler/problems/322_binomial_coefficients_divisible_by_10.txt b/project_euler/problems/322_binomial_coefficients_divisible_by_10.txt index 16fc8a3..ad76d03 100644 --- a/project_euler/problems/322_binomial_coefficients_divisible_by_10.txt +++ b/project_euler/problems/322_binomial_coefficients_divisible_by_10.txt @@ -2,11 +2,11 @@ http://projecteuler.net/problem=322 Binomial coefficients divisible by 10 - -Let T(m, n) be the number of the binomial coefficients iCn that are divisible by 10 for n i m(i, m and n are positive integers). -You are given that T(109, 107-10) = 989697000. - -Find T(1018, 1012-10). +Let T(m, n) be the number of the binomial coefficients iCn that are divisible by 10 for n i m(i, m and n are positive integers). +You are given that T(109, 107-10) = 989697000. + + +Find T(1018, 1012-10). diff --git a/project_euler/problems/323_bitwiseor_operations_on_random_integers.txt b/project_euler/problems/323_bitwiseor_operations_on_random_integers.txt index f0f6cca..242f5e2 100644 --- a/project_euler/problems/323_bitwiseor_operations_on_random_integers.txt +++ b/project_euler/problems/323_bitwiseor_operations_on_random_integers.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=323 Bitwise-OR operations on random integers -Let y0, y1, y2,... be a sequence of random unsigned 32 bit integers +Let y0, y1, y2,... be a sequence of random unsigned 32 bit integers (i.e. 0 yi 232, every value equally likely). For the sequence xi the following recursion is given: @@ -10,6 +10,6 @@ x0 = 0 and xi = xi-1 | yi-1, for i 0. ( | is the bitwise-OR operator) It can be seen that eventually there will be an index N such that xi = 232 -1 (a bit-pattern of all ones) for all i N. -Find the expected value of N. +Find the expected value of N. Give your answer rounded to 10 digits after the decimal point. diff --git a/project_euler/problems/325_stone_game_ii.txt b/project_euler/problems/325_stone_game_ii.txt index bbbafbc..afa387e 100644 --- a/project_euler/problems/325_stone_game_ii.txt +++ b/project_euler/problems/325_stone_game_ii.txt @@ -2,25 +2,25 @@ http://projecteuler.net/problem=325 Stone Game II - -A game is played with two piles of stones and two players. At her turn, a player removes a number of stones from the larger pile. The number of stones she removes must be a positive multiple of the number of stones in the smaller pile. - -E.g., let the ordered pair(6,14) describe a configuration with 6 stones in the smaller pile and 14 stones in the larger pile, then the first player can remove 6 or 12 stones from the larger pile. +A game is played with two piles of stones and two players. At her turn, a player removes a number of stones from the larger pile. The number of stones she removes must be a positive multiple of the number of stones in the smaller pile. - -The player taking all the stones from a pile wins the game. - -A winning configuration is one where the first player can force a win. For example, (1,5), (2,6) and (3,12) are winning configurations because the first player can immediately remove all stones in the second pile. +E.g., let the ordered pair(6,14) describe a configuration with 6 stones in the smaller pile and 14 stones in the larger pile, then the first player can remove 6 or 12 stones from the larger pile. - -A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (2,3) and (3,4) are losing configurations: any legal move leaves a winning configuration for the second player. - -Define S(N) as the sum of (xi+yi) for all losing configurations (xi,yi), 0 xi yi N. We can verify that S(10) = 211 and S(104) = 230312207313. +The player taking all the stones from a pile wins the game. - -Find S(1016) mod 710. + +A winning configuration is one where the first player can force a win. For example, (1,5), (2,6) and (3,12) are winning configurations because the first player can immediately remove all stones in the second pile. + + +A losing configuration is one where the second player can force a win, no matter what the first player does. For example, (2,3) and (3,4) are losing configurations: any legal move leaves a winning configuration for the second player. + + +Define S(N) as the sum of (xi+yi) for all losing configurations (xi,yi), 0 xi yi N. We can verify that S(10) = 211 and S(104) = 230312207313. + + +Find S(1016) mod 710. diff --git a/project_euler/problems/326_modulo_summations.txt b/project_euler/problems/326_modulo_summations.txt index ebffa80..d109c10 100644 --- a/project_euler/problems/326_modulo_summations.txt +++ b/project_euler/problems/326_modulo_summations.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=326 Modulo Summations - -Let an be a sequence recursively defined by: . - -So the first 10 elements of an are: 1,1,0,3,0,3,5,4,1,9. +Let an be a sequence recursively defined by: . + + +So the first 10 elements of an are: 1,1,0,3,0,3,5,4,1,9. + +Let f(N,M) represent the number of pairs (p,q) such that: + +It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and (9,10). -Let f(N,M) represent the number of pairs (p,q) such that: - -It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and (9,10). - You are also given that f(104,103)=97158. - -Find f(1012,106). + +Find f(1012,106). diff --git a/project_euler/problems/328_lowestcost_search.txt b/project_euler/problems/328_lowestcost_search.txt index b201f92..af02a40 100644 --- a/project_euler/problems/328_lowestcost_search.txt +++ b/project_euler/problems/328_lowestcost_search.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=328 Lowest-cost Search -We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions. +We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions. Each number (question) we ask, has a cost equal to the number asked and we get one of three possible answers: "Your guess is lower than the hidden number", or @@ -11,16 +11,16 @@ Each number (question) we ask, has a cost equal to the number asked and we get o Given the value of n, an optimal strategy minimizes the total cost (i.e. the sum of all the questions asked) for the worst possible case. E.g. If n=3, the best we can do is obviously to ask the number "2". The answer will immediately lead us to find the hidden number (at a total cost = 2). -If n=8, we might decide to use a "binary search" type of strategy: Our first question would be "4" and if the hidden number is higher than 4 we will need one or two additional questions. -Let our second question be "6". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8. +If n=8, we might decide to use a "binary search" type of strategy: Our first question would be "4" and if the hidden number is higher than 4 we will need one or two additional questions. +Let our second question be "6". If the hidden number is still higher than 6, we will need a third question in order to discriminate between 7 and 8. Thus, our third question will be "7" and the total cost for this worst-case scenario will be 4+6+7=17. -We can improve considerably the worst-case cost for n=8, by asking "5" as our first question. -If we are told that the hidden number is higher than 5, our second question will be "7", then we'll know for certain what the hidden number is (for a total cost of 5+7=12). -If we are told that the hidden number is lower than 5, our second question will be "3" and if the hidden number is lower than 3 our third question will be "1", giving a total cost of 5+3+1=9. -Since 12>9, the worst-case cost for this strategy is 12. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy. +We can improve considerably the worst-case cost for n=8, by asking "5" as our first question. +If we are told that the hidden number is higher than 5, our second question will be "7", then we'll know for certain what the hidden number is (for a total cost of 5+7=12). +If we are told that the hidden number is lower than 5, our second question will be "3" and if the hidden number is lower than 3 our third question will be "1", giving a total cost of 5+3+1=9. +Since 12>9, the worst-case cost for this strategy is 12. That's better than what we achieved previously with the "binary search" strategy; it is also better than or equal to any other strategy. So, in fact, we have just described an optimal strategy for n=8. -Let C(n) be the worst-case cost achieved by an optimal strategy for n, as described above. -Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12. +Let C(n) be the worst-case cost achieved by an optimal strategy for n, as described above. +Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12. Similarly, C(100) = 400 and C(n) = 17575. Find C(n). diff --git a/project_euler/problems/329_prime_frog.txt b/project_euler/problems/329_prime_frog.txt index 74534f7..a0f9f0e 100644 --- a/project_euler/problems/329_prime_frog.txt +++ b/project_euler/problems/329_prime_frog.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=329 Prime Frog -Susan has a prime frog. -Her frog is jumping around over 500 squares numbered 1 to 500. -He can only jump one square to the left or to the right, with equal probability, and he cannot jump outside the range [1;500].(if it lands at either end, it automatically jumps to the only available square on the next move.) - - -When he is on a square with a prime number on it, he croaks 'P' (PRIME) with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before jumping to the next square. -When he is on a square with a number on it that is not a prime he croaks 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping to the next square. - - -Given that the frog's starting position is random with the same probability for every square, and given that she listens to his first 15 croaks, what is the probability that she hears the sequence PPPPNNPPPNPPNPN? - -Give your answer as a fraction p/q in reduced form. - +Susan has a prime frog. +Her frog is jumping around over 500 squares numbered 1 to 500. +He can only jump one square to the left or to the right, with equal probability, and he cannot jump outside the range [1;500].(if it lands at either end, it automatically jumps to the only available square on the next move.) + + +When he is on a square with a prime number on it, he croaks 'P' (PRIME) with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before jumping to the next square. +When he is on a square with a number on it that is not a prime he croaks 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping to the next square. + + +Given that the frog's starting position is random with the same probability for every square, and given that she listens to his first 15 croaks, what is the probability that she hears the sequence PPPPNNPPPNPPNPN? + +Give your answer as a fraction p/q in reduced form. + diff --git a/project_euler/problems/330_eulers_number.txt b/project_euler/problems/330_eulers_number.txt index 3f5e761..94d698b 100644 --- a/project_euler/problems/330_eulers_number.txt +++ b/project_euler/problems/330_eulers_number.txt @@ -1,12 +1,12 @@ http://projecteuler.net/problem=330 Euler's Number - -An infinite sequence of real numbers a(n) is defined for all integers n as follows: + +An infinite sequence of real numbers a(n) is defined for all integers n as follows: For example, -a(0) = +a(0) = 1 1! @@ -24,10 +24,10 @@ a(0) = 3! -+ ... = e 1 ++ ... = e 1 -a(1) = +a(1) = e 1 1! @@ -45,10 +45,10 @@ e 1 3! -+ ... = 2e 3 ++ ... = 2e 3 -a(2) = +a(2) = 2e 3 1! @@ -72,25 +72,25 @@ e 1 2 -e 6 - -with e = 2.7182818... being Euler's constant. - +e 6 + +with e = 2.7182818... being Euler's constant. + + +It can be shown that a(n) is of the form -It can be shown that a(n) is of the form - A(n) e + B(n) n! -for integers A(n) and B(n). - +for integers A(n) and B(n). -For example a(10) = - + +For example a(10) = + 328161643 e 652694486 10! @@ -98,7 +98,7 @@ For example a(10) = . - -Find A(109) + B(109) and give your answer mod 77 777 777. + +Find A(109) + B(109) and give your answer mod 77 777 777. diff --git a/project_euler/problems/331_cross_flips.txt b/project_euler/problems/331_cross_flips.txt index b31ae47..90168e6 100644 --- a/project_euler/problems/331_cross_flips.txt +++ b/project_euler/problems/331_cross_flips.txt @@ -6,11 +6,11 @@ NN disks are placed on a square game board. Each disk has a black side and white At each turn, you may choose a disk and flip all the disks in the same row and the same column as this disk: thus 2N-1 disks are flipped. The game ends when all disks show their white side. The following example shows a game on a 55 board. It can be proven that 3 is the minimal number of turns to finish this game. -The bottom left disk on the NN board has coordinates (0,0); +The bottom left disk on the NN board has coordinates (0,0); the bottom right disk has coordinates (N-1,0) and the top left disk has coordinates (0,N-1). -Let CN be the following configuration of a board with NN disks: +Let CN be the following configuration of a board with NN disks: A disk at (x,y) satisfying , shows its black side; otherwise, it shows its white side. C5 is shown above. -Let T(N) be the minimal number of turns to finish a game starting from configuration CN or 0 if configuration CN is unsolvable. +Let T(N) be the minimal number of turns to finish a game starting from configuration CN or 0 if configuration CN is unsolvable. We have shown that T(5)=3. You are also given that T(10)=29 and T(1 000)=395253. Find . diff --git a/project_euler/problems/332_spherical_triangles.txt b/project_euler/problems/332_spherical_triangles.txt index 0428a29..e9cdad7 100644 --- a/project_euler/problems/332_spherical_triangles.txt +++ b/project_euler/problems/332_spherical_triangles.txt @@ -4,10 +4,10 @@ Spherical triangles A spherical triangle is a figure formed on the surface of a sphere by three great circular arcs intersecting pairwise in three vertices. -Let C(r) be the sphere with the centre (0,0,0) and radius r. -Let Z(r) be the set of points on the surface of C(r) with integer coordinates. -Let T(r) be the set of spherical triangles with vertices in Z(r). -Degenerate spherical triangles, formed by three points on the same great arc, are not included in T(r). +Let C(r) be the sphere with the centre (0,0,0) and radius r. +Let Z(r) be the set of points on the surface of C(r) with integer coordinates. +Let T(r) be the set of spherical triangles with vertices in Z(r). +Degenerate spherical triangles, formed by three points on the same great arc, are not included in T(r). Let A(r) be the area of the smallest spherical triangle in T(r). For example A(14) is 3.294040 rounded to six decimal places. Find A(r). Give your answer rounded to six decimal places. diff --git a/project_euler/problems/333_special_partitions.txt b/project_euler/problems/333_special_partitions.txt index c2480ef..830b12c 100644 --- a/project_euler/problems/333_special_partitions.txt +++ b/project_euler/problems/333_special_partitions.txt @@ -3,10 +3,10 @@ http://projecteuler.net/problem=333 Special partitions All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as 2ix3j, where i,j 0. -Let's consider only those such partitions where none of the terms can divide any of the other terms. +Let's consider only those such partitions where none of the terms can divide any of the other terms. For example, the partition of 17 = 2 + 6 + 9 = (21x30 + 21x31 + 20x32) would not be valid since 2 can divide 6. Neither would the partition 17 = 16 + 1 = (24x30 + 20x30) since 1 can divide 16. The only valid partition of 17 would be 8 + 9 = (23x30 + 20x32). -Many integers have more than one valid partition, the first being 11 having the following two partitions. -11 = 2 + 9 = (21x30 + 20x32) +Many integers have more than one valid partition, the first being 11 having the following two partitions. +11 = 2 + 9 = (21x30 + 20x32) 11 = 8 + 3 = (23x30 + 20x31) Let's define P(n) as the number of valid partitions of n. For example, P(11) = 2. Let's consider only the prime integers q which would have a single valid partition such as P(17). diff --git a/project_euler/problems/334_spilling_the_beans.txt b/project_euler/problems/334_spilling_the_beans.txt index d3e2881..523368e 100644 --- a/project_euler/problems/334_spilling_the_beans.txt +++ b/project_euler/problems/334_spilling_the_beans.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=334 Spilling the beans -In Plato's heaven, there exist an infinite number of bowls in a straight line. -Each bowl either contains some or none of a finite number of beans. +In Plato's heaven, there exist an infinite number of bowls in a straight line. +Each bowl either contains some or none of a finite number of beans. A child plays a game, which allows only one kind of move: removing two beans from any bowl, and putting one in each of the two adjacent bowls. The game ends when each bowl contains either one or no beans. For example, consider two adjacent bowls containing 2 and 3 beans respectively, all other bowls being empty. The following eight moves will finish the game: @@ -11,14 +11,14 @@ You are given the following sequences: -t0 = 123456. - +t0 = 123456. -ti = - + +ti = + @@ -30,13 +30,13 @@ ti-1 2 - - , - - - if ti-1 is even - + , + + + + if ti-1 is even + @@ -49,37 +49,37 @@ ti-1 - - 926252, - - - if ti-1 is odd - + + 926252, + + + if ti-1 is odd + + + + + + + where x is the floor function - - where x is the floor function - + and is the bitwise XOR operator. - - and is the bitwise XOR operator. - +bi = ( ti mod 211) + 1. -bi = ( ti mod 211) + 1. - -The first two terms of the last sequence are b1 = 289 and b2 = 145. +The first two terms of the last sequence are b1 = 289 and b2 = 145. If we start with b1 and b2 beans in two adjacent bowls, 3419100 moves would be required to finish the game. Consider now 1500 adjacent bowls containing b1, b2,..., b1500 beans respectively, all other bowls being empty. Find how many moves it takes before the game ends. diff --git a/project_euler/problems/336_maximix_arrangements.txt b/project_euler/problems/336_maximix_arrangements.txt index b112d90..a2ec642 100644 --- a/project_euler/problems/336_maximix_arrangements.txt +++ b/project_euler/problems/336_maximix_arrangements.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=336 Maximix Arrangements -A train is used to transport four carriages in the order: ABCD. However, sometimes when the train arrives to collect the carriages they are not in the correct order. -To rearrange the carriages they are all shunted on to a large rotating turntable. After the carriages are uncoupled at a specific point the train moves off the turntable pulling the carriages still attached with it. The remaining carriages are rotated 180 degrees. All of the carriages are then rejoined and this process is repeated as often as necessary in order to obtain the least number of uses of the turntable. +A train is used to transport four carriages in the order: ABCD. However, sometimes when the train arrives to collect the carriages they are not in the correct order. +To rearrange the carriages they are all shunted on to a large rotating turntable. After the carriages are uncoupled at a specific point the train moves off the turntable pulling the carriages still attached with it. The remaining carriages are rotated 180 degrees. All of the carriages are then rejoined and this process is repeated as often as necessary in order to obtain the least number of uses of the turntable. Some arrangements, such as ADCB, can be solved easily: the carriages are separated between A and D, and after DCB are rotated the correct order has been achieved. However, Simple Simon, the train driver, is not known for his efficiency, so he always solves the problem by initially getting carriage A in the correct place, then carriage B, and so on. Using four carriages, the worst possible arrangements for Simon, which we shall call maximix arrangements, are DACB and DBAC; each requiring him five rotations (although, using the most efficient approach, they could be solved using just three rotations). The process he uses for DACB is shown below. diff --git a/project_euler/problems/337_totient_stairstep_sequences.txt b/project_euler/problems/337_totient_stairstep_sequences.txt index bec0c22..fd58050 100644 --- a/project_euler/problems/337_totient_stairstep_sequences.txt +++ b/project_euler/problems/337_totient_stairstep_sequences.txt @@ -7,8 +7,8 @@ Let {a1, a2,..., an} be an integer sequence of length n such that: a1 = 6 for all 1 i n : φ(ai) i+1) i i+1 1 -Let S(N) be the number of such sequences with an N. -For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}. +Let S(N) be the number of such sequences with an N. +For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}. We can verify that S(100) = 482073668 and S(10 000) mod 108 = 73808307. Find S(20 000 000) mod 108. 1 φ denotes Euler's totient function. diff --git a/project_euler/problems/338_cutting_rectangular_grid_paper.txt b/project_euler/problems/338_cutting_rectangular_grid_paper.txt index 552e378..2bef7fb 100644 --- a/project_euler/problems/338_cutting_rectangular_grid_paper.txt +++ b/project_euler/problems/338_cutting_rectangular_grid_paper.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=338 Cutting Rectangular Grid Paper -A rectangular sheet of grid paper with integer dimensions w h is given. Its grid spacing is 1. +A rectangular sheet of grid paper with integer dimensions w h is given. Its grid spacing is 1. When we cut the sheet along the grid lines into two pieces and rearrange those pieces without overlap, we can make new rectangles with different dimensions. For example, from a sheet with dimensions 9 4 , we can make rectangles with dimensions 18 2, 12 3 and 6 6 by cutting and rearranging as below: Similarly, from a sheet with dimensions 9 8 , we can make rectangles with dimensions 18 4 and 12 6 . -For a pair w and h, let F(w,h) be the number of distinct rectangles that can be made from a sheet with dimensions w h . -For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2. -Note that rectangles congruent to the initial one are not counted in F(w,h). +For a pair w and h, let F(w,h) be the number of distinct rectangles that can be made from a sheet with dimensions w h . +For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2. +Note that rectangles congruent to the initial one are not counted in F(w,h). Note also that rectangles with dimensions w h and dimensions h w are not considered distinct. -For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h which satisfy 0 h w N. +For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h which satisfy 0 h w N. We can verify that G(10) = 55, G(103) = 971745 and G(105) = 9992617687. Find G(1012). Give your answer modulo 108. diff --git a/project_euler/problems/339_peredur_fab_efrawg.txt b/project_euler/problems/339_peredur_fab_efrawg.txt index bac3aa0..d4bd559 100644 --- a/project_euler/problems/339_peredur_fab_efrawg.txt +++ b/project_euler/problems/339_peredur_fab_efrawg.txt @@ -6,13 +6,13 @@ Peredur fab Efrawg "And he came towards a valley, through which ran a river; and the borders of the valley were wooded, and on each side of the river were level meadows. And on one side of the river he saw a flock of white sheep, and on the other a flock of black sheep. And whenever one of the white sheep bleated, one of the black sheep would cross over and become white; and when one of the black sheep bleated, one of the white sheep would cross over and become black." en.wikisource.org - - -Initially each flock consists of n sheep. Each sheep (regardless of colour) is equally likely to be the next sheep to bleat. After a sheep has bleated and a sheep from the other flock has crossed over, Peredur may remove a number of white sheep in order to maximize the expected final number of black sheep. Let E(n) be the expected final number of black sheep if Peredur uses an optimal strategy. - - - -You are given that E(5) = 6.871346 rounded to 6 places behind the decimal point. -Find E(10 000) and give your answer rounded to 6 places behind the decimal point. + + +Initially each flock consists of n sheep. Each sheep (regardless of colour) is equally likely to be the next sheep to bleat. After a sheep has bleated and a sheep from the other flock has crossed over, Peredur may remove a number of white sheep in order to maximize the expected final number of black sheep. Let E(n) be the expected final number of black sheep if Peredur uses an optimal strategy. + + + +You are given that E(5) = 6.871346 rounded to 6 places behind the decimal point. +Find E(10 000) and give your answer rounded to 6 places behind the decimal point. diff --git a/project_euler/problems/340_crazy_function.txt b/project_euler/problems/340_crazy_function.txt index 2d872ae..c629608 100644 --- a/project_euler/problems/340_crazy_function.txt +++ b/project_euler/problems/340_crazy_function.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=340 Crazy Function - -For fixed integers a, b, c, define the crazy function F(n) as follows: -F(n) = n - c for all n b -F(n) = F(a + F(a + F(a + F(a + n)))) for all n b. - -Also, define S(a, b, c) = . +For fixed integers a, b, c, define the crazy function F(n) as follows: +F(n) = n - c for all n b +F(n) = F(a + F(a + F(a + F(a + n)))) for all n b. - -For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000) = 2040. -Also, S(50, 2000, 40) = 5204240. - -Find the last 9 digits of S(217, 721, 127). +Also, define S(a, b, c) = . + + +For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000) = 2040. +Also, S(50, 2000, 40) = 5204240. + + +Find the last 9 digits of S(217, 721, 127). diff --git a/project_euler/problems/341_golombs_selfdescribing_sequence.txt b/project_euler/problems/341_golombs_selfdescribing_sequence.txt index eb6da4a..3b0f453 100644 --- a/project_euler/problems/341_golombs_selfdescribing_sequence.txt +++ b/project_euler/problems/341_golombs_selfdescribing_sequence.txt @@ -7,7 +7,7 @@ The Golomb's self-describing sequence {G(n)} is the only nondecreasing sequence n123456789101112131415… G(n)122334445556666… -You are given that G(103) = 86, G(106) = 6137. +You are given that G(103) = 86, G(106) = 6137. You are also given that ΣG(n3) = 153506976 for 1 n 103. Find ΣG(n3) for 1 n 106. diff --git a/project_euler/problems/342_the_totient_of_a_square_is_a_cube.txt b/project_euler/problems/342_the_totient_of_a_square_is_a_cube.txt index bd74735..6abf8b7 100644 --- a/project_euler/problems/342_the_totient_of_a_square_is_a_cube.txt +++ b/project_euler/problems/342_the_totient_of_a_square_is_a_cube.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=342 The totient of a square is a cube - -Consider the number 50. -502 = 2500 = 22 54, so φ(2500) = 2 4 53 = 8 53 = 23 53. 1 -So 2500 is a square and φ(2500) is a cube. - -Find the sum of all numbers n, 1 < n 1010 such that φ(n2) is a cube. +Consider the number 50. +502 = 2500 = 22 54, so φ(2500) = 2 4 53 = 8 53 = 23 53. 1 +So 2500 is a square and φ(2500) is a cube. -1 φ denotes Euler's totient function. +Find the sum of all numbers n, 1 < n 1010 such that φ(n2) is a cube. + + +1 φ denotes Euler's totient function. diff --git a/project_euler/problems/343_fractional_sequences.txt b/project_euler/problems/343_fractional_sequences.txt index c282e16..49e899d 100644 --- a/project_euler/problems/343_fractional_sequences.txt +++ b/project_euler/problems/343_fractional_sequences.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=343 Fractional Sequences -For any positive integer k, a finite sequence ai of fractions xi/yi is defined by: -a1 = 1/k and -ai = (xi-1+1)/(yi-1-1) reduced to lowest terms for i>1. -When ai reaches some integer n, the sequence stops. (That is, when yi=1.) -Define f(k) = n. -For example, for k = 20: +For any positive integer k, a finite sequence ai of fractions xi/yi is defined by: +a1 = 1/k and +ai = (xi-1+1)/(yi-1-1) reduced to lowest terms for i>1. +When ai reaches some integer n, the sequence stops. (That is, when yi=1.) +Define f(k) = n. +For example, for k = 20: - -1/20 2/19 3/18 = 1/6 2/5 3/4 4/3 5/2 6/1 = 6 - -So f(20) = 6. +1/20 2/19 3/18 = 1/6 2/5 3/4 4/3 5/2 6/1 = 6 - -Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k3) = 118937 for 1 k 100. - -Find Σf(k3) for 1 k 2106. +So f(20) = 6. + + +Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k3) = 118937 for 1 k 100. + + +Find Σf(k3) for 1 k 2106. diff --git a/project_euler/problems/344_silver_dollar_game.txt b/project_euler/problems/344_silver_dollar_game.txt index aa4888b..9eab01d 100644 --- a/project_euler/problems/344_silver_dollar_game.txt +++ b/project_euler/problems/344_silver_dollar_game.txt @@ -13,6 +13,6 @@ The winner is the player who pockets the silver dollar. A winning configuration is an arrangement of coins on the strip where the first player can force a win no matter what the second player does. Let W(n,c) be the number of winning configurations for a strip of n squares, c worthless coins and one silver dollar. You are given that W(10,2) = 324 and W(100,10) = 1514704946113500. -Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003 · 1 000 033). +Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003 · 1 000 033). diff --git a/project_euler/problems/345_matrix_sum.txt b/project_euler/problems/345_matrix_sum.txt index dedea5a..42e4f83 100644 --- a/project_euler/problems/345_matrix_sum.txt +++ b/project_euler/problems/345_matrix_sum.txt @@ -3,30 +3,30 @@ http://projecteuler.net/problem=345 Matrix Sum We define the Matrix Sum of a matrix as the maximum sum of matrix elements with each element being the only one in his row and column. For example, the Matrix Sum of the matrix below equals 3315 ( = 863 + 383 + 343 + 959 + 767): - -  7  53 183 439 863 -497 383 563  79 973 -287  63 343 169 583 + +  7  53 183 439 863 +497 383 563  79 973 +287  63 343 169 583 627 343 773 959 943 767 473 103 699 303 - + Find the Matrix Sum of: - -  7  53 183 439 863 497 383 563  79 973 287  63 343 169 583 -627 343 773 959 943 767 473 103 699 303 957 703 583 639 913 -447 283 463  29  23 487 463 993 119 883 327 493 423 159 743 -217 623   3 399 853 407 103 983  89 463 290 516 212 462 350 -960 376 682 962 300 780 486 502 912 800 250 346 172 812 350 -870 456 192 162 593 473 915  45 989 873 823 965 425 329 803 -973 965 905 919 133 673 665 235 509 613 673 815 165 992 326 -322 148 972 962 286 255 941 541 265 323 925 281 601  95 973 -445 721  11 525 473  65 511 164 138 672  18 428 154 448 848 -414 456 310 312 798 104 566 520 302 248 694 976 430 392 198 -184 829 373 181 631 101 969 613 840 740 778 458 284 760 390 -821 461 843 513  17 901 711 993 293 157 274  94 192 156 574 - 34 124   4 878 450 476 712 914 838 669 875 299 823 329 699 -815 559 813 459 522 788 168 586 966 232 308 833 251 631 107 + +  7  53 183 439 863 497 383 563  79 973 287  63 343 169 583 +627 343 773 959 943 767 473 103 699 303 957 703 583 639 913 +447 283 463  29  23 487 463 993 119 883 327 493 423 159 743 +217 623   3 399 853 407 103 983  89 463 290 516 212 462 350 +960 376 682 962 300 780 486 502 912 800 250 346 172 812 350 +870 456 192 162 593 473 915  45 989 873 823 965 425 329 803 +973 965 905 919 133 673 665 235 509 613 673 815 165 992 326 +322 148 972 962 286 255 941 541 265 323 925 281 601  95 973 +445 721  11 525 473  65 511 164 138 672  18 428 154 448 848 +414 456 310 312 798 104 566 520 302 248 694 976 430 392 198 +184 829 373 181 631 101 969 613 840 740 778 458 284 760 390 +821 461 843 513  17 901 711 993 293 157 274  94 192 156 574 + 34 124   4 878 450 476 712 914 838 669 875 299 823 329 699 +815 559 813 459 522 788 168 586 966 232 308 833 251 631 107 813 883 451 509 615  77 281 613 459 205 380 274 302  35 805 diff --git a/project_euler/problems/346_strong_repunits.txt b/project_euler/problems/346_strong_repunits.txt index f549abd..d3d59d7 100644 --- a/project_euler/problems/346_strong_repunits.txt +++ b/project_euler/problems/346_strong_repunits.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=346 Strong Repunits - -The number 7 is special, because 7 is 111 written in base 2, and 11 written in base 6 (i.e. 710 = 116 = 1112). In other words, 7 is a repunit in at least two bases b > 1. - - -We shall call a positive integer with this property a strong repunit. It can be verified that there are 8 strong repunits below 50: {1,7,13,15,21,31,40,43}. Furthermore, the sum of all strong repunits below 1000 equals 15864. - -Find the sum of all strong repunits below 1012. - - - + +The number 7 is special, because 7 is 111 written in base 2, and 11 written in base 6 (i.e. 710 = 116 = 1112). In other words, 7 is a repunit in at least two bases b > 1. + + +We shall call a positive integer with this property a strong repunit. It can be verified that there are 8 strong repunits below 50: {1,7,13,15,21,31,40,43}. Furthermore, the sum of all strong repunits below 1000 equals 15864. + +Find the sum of all strong repunits below 1012. + + + diff --git a/project_euler/problems/347_largest_integer_divisible_by_two_primes.txt b/project_euler/problems/347_largest_integer_divisible_by_two_primes.txt index 4c214c5..fa6f428 100644 --- a/project_euler/problems/347_largest_integer_divisible_by_two_primes.txt +++ b/project_euler/problems/347_largest_integer_divisible_by_two_primes.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=347 Largest integer divisible by two primes - -The largest integer 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32*3=25*3. -For two distinct primes p and q let M(p,q,N) be the largest positive integer N only divisible -by both p and q and M(p,q,N)=0 if such a positive integer does not exist. - - -E.g. M(2,3,100)=96. -M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. -Also M(2,73,100)=0 because there does not exist a positive integer 100 that is divisible by both 2 and 73. - - -Let S(N) be the sum of all distinct M(p,q,N). -S(100)=2262. - - -Find S(10 000 000). + +The largest integer 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32*3=25*3. +For two distinct primes p and q let M(p,q,N) be the largest positive integer N only divisible +by both p and q and M(p,q,N)=0 if such a positive integer does not exist. + + +E.g. M(2,3,100)=96. +M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. +Also M(2,73,100)=0 because there does not exist a positive integer 100 that is divisible by both 2 and 73. + + +Let S(N) be the sum of all distinct M(p,q,N). +S(100)=2262. + + +Find S(10 000 000). diff --git a/project_euler/problems/348_sum_of_a_square_and_a_cube.txt b/project_euler/problems/348_sum_of_a_square_and_a_cube.txt index e178b90..5d8bae9 100644 --- a/project_euler/problems/348_sum_of_a_square_and_a_cube.txt +++ b/project_euler/problems/348_sum_of_a_square_and_a_cube.txt @@ -3,11 +3,11 @@ http://projecteuler.net/problem=348 Sum of a square and a cube Many numbers can be expressed as the sum of a square and a cube. Some of them in more than one way. -Consider the palindromic numbers that can be expressed as the sum of a square and a cube, both greater than 1, in exactly 4 different ways. +Consider the palindromic numbers that can be expressed as the sum of a square and a cube, both greater than 1, in exactly 4 different ways. For example, 5229225 is a palindromic number and it can be expressed in exactly 4 different ways: -22852 + 203 -22232 + 663 -18102 + 1253 +22852 + 203 +22232 + 663 +18102 + 1253 11972 + 1563 Find the sum of the five smallest such palindromic numbers. diff --git a/project_euler/problems/349_langtons_ant.txt b/project_euler/problems/349_langtons_ant.txt index 4254063..445dfe0 100644 --- a/project_euler/problems/349_langtons_ant.txt +++ b/project_euler/problems/349_langtons_ant.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=349 Langton's ant - -An ant moves on a regular grid of squares that are coloured either black or white. -The ant is always oriented in one of the cardinal directions (left, right, up or down) and moves from square to adjacent square according to the following rules: -- if it is on a black square, it flips the color of the square to white, rotates 90 degrees counterclockwise and moves forward one square. + +An ant moves on a regular grid of squares that are coloured either black or white. +The ant is always oriented in one of the cardinal directions (left, right, up or down) and moves from square to adjacent square according to the following rules: +- if it is on a black square, it flips the color of the square to white, rotates 90 degrees counterclockwise and moves forward one square. - if it is on a white square, it flips the color of the square to black, rotates 90 degrees clockwise and moves forward one square. - -Starting with a grid that is entirely white, how many squares are black after 1018 moves of the ant? + +Starting with a grid that is entirely white, how many squares are black after 1018 moves of the ant? diff --git a/project_euler/problems/350_constraining_the_least_greatest_and_the_greatest_least.txt b/project_euler/problems/350_constraining_the_least_greatest_and_the_greatest_least.txt index b4d152a..b38c472 100644 --- a/project_euler/problems/350_constraining_the_least_greatest_and_the_greatest_least.txt +++ b/project_euler/problems/350_constraining_the_least_greatest_and_the_greatest_least.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=350 Constraining the least greatest and the greatest least -A list of size n is a sequence of n natural numbers. Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11). - -The greatest common divisor, or gcd, of a list is the largest natural number that divides all entries of the list. Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11. - -The least common multiple, or lcm, of a list is the smallest natural number divisible by each entry of the list. Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11. - -Let f(G, L, N) be the number of lists of size N with gcd G and lcm L. For example: - -f(10, 100, 1) = 91. -f(10, 100, 2) = 327. -f(10, 100, 3) = 1135. -f(10, 100, 1000) mod 1014 = 3286053. - -Find f(106, 1012, 1018) mod 1014. +A list of size n is a sequence of n natural numbers. Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11). + +The greatest common divisor, or gcd, of a list is the largest natural number that divides all entries of the list. Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11. + +The least common multiple, or lcm, of a list is the smallest natural number divisible by each entry of the list. Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11. + +Let f(G, L, N) be the number of lists of size N with gcd G and lcm L. For example: + +f(10, 100, 1) = 91. +f(10, 100, 2) = 327. +f(10, 100, 3) = 1135. +f(10, 100, 1000) mod 1014 = 3286053. + +Find f(106, 1012, 1018) mod 1014. diff --git a/project_euler/problems/351_hexagonal_orchards.txt b/project_euler/problems/351_hexagonal_orchards.txt index 114822e..daba293 100644 --- a/project_euler/problems/351_hexagonal_orchards.txt +++ b/project_euler/problems/351_hexagonal_orchards.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=351 Hexagonal orchards -A hexagonal orchard of order n is a triangular lattice made up of points within a regular hexagon with side n. The following is an example of a hexagonal orchard of order 5: +A hexagonal orchard of order n is a triangular lattice made up of points within a regular hexagon with side n. The following is an example of a hexagonal orchard of order 5: - -Highlighted in green are the points which are hidden from the center by a point closer to it. It can be seen that for a hexagonal orchard of order 5, 30 points are hidden from the center. - -Let H(n) be the number of points hidden from the center in a hexagonal orchard of order n. +Highlighted in green are the points which are hidden from the center by a point closer to it. It can be seen that for a hexagonal orchard of order 5, 30 points are hidden from the center. - -H(5) = 30. H(10) = 138. H(1 000) = 1177848. - -Find H(100 000 000). +Let H(n) be the number of points hidden from the center in a hexagonal orchard of order n. + + +H(5) = 30. H(10) = 138. H(1 000) = 1177848. + + +Find H(100 000 000). diff --git a/project_euler/problems/352_blood_tests.txt b/project_euler/problems/352_blood_tests.txt index d7d2676..7a2fe30 100644 --- a/project_euler/problems/352_blood_tests.txt +++ b/project_euler/problems/352_blood_tests.txt @@ -2,47 +2,47 @@ http://projecteuler.net/problem=352 Blood tests - -Each one of the 25 sheep in a flock must be tested for a rare virus, known to affect 2% of the sheep population. -An accurate and extremely sensitive PCR test exists for blood samples, producing a clear positive / negative result, but it is very time-consuming and expensive. - -Because of the high cost, the vet-in-charge suggests that instead of performing 25 separate tests, the following procedure can be used instead: -The sheep are split into 5 groups of 5 sheep in each group. -For each group, the 5 samples are mixed together and a single test is performed. Then, +Each one of the 25 sheep in a flock must be tested for a rare virus, known to affect 2% of the sheep population. +An accurate and extremely sensitive PCR test exists for blood samples, producing a clear positive / negative result, but it is very time-consuming and expensive. + + +Because of the high cost, the vet-in-charge suggests that instead of performing 25 separate tests, the following procedure can be used instead: +The sheep are split into 5 groups of 5 sheep in each group. +For each group, the 5 samples are mixed together and a single test is performed. Then, If the result is negative, all the sheep in that group are deemed to be virus-free. If the result is positive, 5 additional tests will be performed (a separate test for each animal) to determine the affected individual(s). - -Since the probability of infection for any specific animal is only 0.02, the first test (on the pooled samples) for each group will be: + +Since the probability of infection for any specific animal is only 0.02, the first test (on the pooled samples) for each group will be: Negative (and no more tests needed) with probability 0.985 = 0.9039207968. Positive (5 additional tests needed) with probability 1 - 0.9039207968 = 0.0960792032. - -Thus, the expected number of tests for each group is 1 + 0.0960792032 5 = 1.480396016. -Consequently, all 5 groups can be screened using an average of only 1.480396016 5 = 7.40198008 tests, which represents a huge saving of more than 70% ! - -Although the scheme we have just described seems to be very efficient, it can still be improved considerably (always assuming that the test is sufficiently sensitive and that there are no adverse effects caused by mixing different samples). E.g.: +Thus, the expected number of tests for each group is 1 + 0.0960792032 5 = 1.480396016. +Consequently, all 5 groups can be screened using an average of only 1.480396016 5 = 7.40198008 tests, which represents a huge saving of more than 70% ! + + +Although the scheme we have just described seems to be very efficient, it can still be improved considerably (always assuming that the test is sufficiently sensitive and that there are no adverse effects caused by mixing different samples). E.g.: We may start by running a test on a mixture of all the 25 samples. It can be verified that in about 60.35% of the cases this test will be negative, thus no more tests will be needed. Further testing will only be required for the remaining 39.65% of the cases. If we know that at least one animal in a group of 5 is infected and the first 4 individual tests come out negative, there is no need to run a test on the fifth animal (we know that it must be infected). We can try a different number of groups / different number of animals in each group, adjusting those numbers at each level so that the total expected number of tests will be minimised. - -To simplify the very wide range of possibilities, there is one restriction we place when devising the most cost-efficient testing scheme: whenever we start with a mixed sample, all the sheep contributing to that sample must be fully screened (i.e. a verdict of infected / virus-free must be reached for all of them) before we start examining any other animals. - -For the current example, it turns out that the most cost-efficient testing scheme (we'll call it the optimal strategy) requires an average of just 4.155452 tests! - - - -Using the optimal strategy, let T(s,p) represent the average number of tests needed to screen a flock of s sheep for a virus having probability p to be present in any individual. -Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25, 0.10) = 12.702124. - - -Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50. -Give your answer rounded to six decimal places. + +To simplify the very wide range of possibilities, there is one restriction we place when devising the most cost-efficient testing scheme: whenever we start with a mixed sample, all the sheep contributing to that sample must be fully screened (i.e. a verdict of infected / virus-free must be reached for all of them) before we start examining any other animals. + +For the current example, it turns out that the most cost-efficient testing scheme (we'll call it the optimal strategy) requires an average of just 4.155452 tests! + + + +Using the optimal strategy, let T(s,p) represent the average number of tests needed to screen a flock of s sheep for a virus having probability p to be present in any individual. +Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25, 0.10) = 12.702124. + + +Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50. +Give your answer rounded to six decimal places. diff --git a/project_euler/problems/353_risky_moon.txt b/project_euler/problems/353_risky_moon.txt index ebeb5ef..4b731d7 100644 --- a/project_euler/problems/353_risky_moon.txt +++ b/project_euler/problems/353_risky_moon.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=353 Risky moon - -A moon could be described by the sphere C(r) with centre (0,0,0) and radius r. - -There are stations on the moon at the points on the surface of C(r) with integer coordinates. The station at (0,0,r) is called North Pole station, the station at (0,0,-r) is called South Pole station. +A moon could be described by the sphere C(r) with centre (0,0,0) and radius r. - -All stations are connected with each other via the shortest road on the great arc through the stations. A journey between two stations is risky. If d is the length of the road between two stations, (d/(π r))2 is a measure for the risk of the journey (let us call it the risk of the road). If the journey includes more than two stations, the risk of the journey is the sum of risks of the used roads. - -A direct journey from the North Pole station to the South Pole station has the length πr and risk 1. The journey from the North Pole station to the South Pole station via (0,r,0) has the same length, but a smaller risk: (½πr/(πr))2+(½πr/(πr))2=0.5. +There are stations on the moon at the points on the surface of C(r) with integer coordinates. The station at (0,0,r) is called North Pole station, the station at (0,0,-r) is called South Pole station. - -The minimal risk of a journey from the North Pole station to the South Pole station on C(r) is M(r). - -You are given that M(7)=0.1784943998 rounded to 10 digits behind the decimal point. +All stations are connected with each other via the shortest road on the great arc through the stations. A journey between two stations is risky. If d is the length of the road between two stations, (d/(π r))2 is a measure for the risk of the journey (let us call it the risk of the road). If the journey includes more than two stations, the risk of the journey is the sum of risks of the used roads. - -Find M(2n-1) for 1n15. - -Give your answer rounded to 10 digits behind the decimal point in the form a.bcdefghijk. +A direct journey from the North Pole station to the South Pole station has the length πr and risk 1. The journey from the North Pole station to the South Pole station via (0,r,0) has the same length, but a smaller risk: (½πr/(πr))2+(½πr/(πr))2=0.5. + + +The minimal risk of a journey from the North Pole station to the South Pole station on C(r) is M(r). + + +You are given that M(7)=0.1784943998 rounded to 10 digits behind the decimal point. + + +Find M(2n-1) for 1n15. + + +Give your answer rounded to 10 digits behind the decimal point in the form a.bcdefghijk. diff --git a/project_euler/problems/354_distances_in_a_bees_honeycomb.txt b/project_euler/problems/354_distances_in_a_bees_honeycomb.txt index b4a88f3..9725771 100644 --- a/project_euler/problems/354_distances_in_a_bees_honeycomb.txt +++ b/project_euler/problems/354_distances_in_a_bees_honeycomb.txt @@ -1,14 +1,14 @@ http://projecteuler.net/problem=354 -Distances in a bee's honeycomb +Distances in a bee's honeycomb Consider a honey bee's honeycomb where each cell is a perfect regular hexagon with side length 1. - -One particular cell is occupied by the queen bee. -For a positive real number L, let B(L) count the cells with distance L from the queen bee cell (all distances are measured from centre to centre); you may assume that the honeycomb is large enough to accommodate for any distance we wish to consider. + +One particular cell is occupied by the queen bee. +For a positive real number L, let B(L) count the cells with distance L from the queen bee cell (all distances are measured from centre to centre); you may assume that the honeycomb is large enough to accommodate for any distance we wish to consider. For example, B(3) = 6, B(21) = 12 and B(111 111 111) = 54. Find the number of L 5·1011 such that B(L) = 450. diff --git a/project_euler/problems/355_maximal_coprime_subset.txt b/project_euler/problems/355_maximal_coprime_subset.txt index c4caecc..f826d0a 100644 --- a/project_euler/problems/355_maximal_coprime_subset.txt +++ b/project_euler/problems/355_maximal_coprime_subset.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=355 Maximal coprime subset - -Define Co(n) to be the maximal possible sum of a set of mutually co-prime elements from {1, 2, ..., n}. For example Co(10) is 30 and hits that maximum on the subset {1, 5, 7, 8, 9}. - -You are given that Co(30) = 193 and Co(100) = 1356. +Define Co(n) to be the maximal possible sum of a set of mutually co-prime elements from {1, 2, ..., n}. For example Co(10) is 30 and hits that maximum on the subset {1, 5, 7, 8, 9}. -Find Co(200000). + +You are given that Co(30) = 193 and Co(100) = 1356. + +Find Co(200000). diff --git a/project_euler/problems/356_largest_roots_of_cubic_polynomials.txt b/project_euler/problems/356_largest_roots_of_cubic_polynomials.txt index 8acb7d8..9841348 100644 --- a/project_euler/problems/356_largest_roots_of_cubic_polynomials.txt +++ b/project_euler/problems/356_largest_roots_of_cubic_polynomials.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=356 Largest roots of cubic polynomials - -Let an be the largest real root of a polynomial g(x) = x3 - 2n·x2 + n. + +Let an be the largest real root of a polynomial g(x) = x3 - 2n·x2 + n. For example, a2 = 3.86619826... - + Find the last eight digits of. Note: represents the floor function. diff --git a/project_euler/problems/357_prime_generating_integers.txt b/project_euler/problems/357_prime_generating_integers.txt index e59c07e..260e276 100644 --- a/project_euler/problems/357_prime_generating_integers.txt +++ b/project_euler/problems/357_prime_generating_integers.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=357 Prime generating integers - -Consider the divisors of 30: 1,2,3,5,6,10,15,30. -It can be seen that for every divisor d of 30, d+30/d is prime. - -Find the sum of all positive integers n not exceeding 100 000 000such that -for every divisor d of n, d+n/d is prime. +Consider the divisors of 30: 1,2,3,5,6,10,15,30. +It can be seen that for every divisor d of 30, d+30/d is prime. + + +Find the sum of all positive integers n not exceeding 100 000 000such that +for every divisor d of n, d+n/d is prime. diff --git a/project_euler/problems/358_cyclic_numbers.txt b/project_euler/problems/358_cyclic_numbers.txt index 22c2fb1..dd63090 100644 --- a/project_euler/problems/358_cyclic_numbers.txt +++ b/project_euler/problems/358_cyclic_numbers.txt @@ -2,30 +2,30 @@ http://projecteuler.net/problem=358 Cyclic numbers -A cyclic number with n digits has a very interesting property: -When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly the same digits, in the same order, but rotated in a circular fashion! - - -The smallest cyclic number is the 6-digit number 142857 : -142857 1 = 142857 -142857 2 = 285714 -142857 3 = 428571 -142857 4 = 571428 -142857 5 = 714285 -142857 6 = 857142 - - -The next cyclic number is 0588235294117647 with 16 digits : -0588235294117647 1 = 0588235294117647 -0588235294117647 2 = 1176470588235294 -0588235294117647 3 = 1764705882352941 -... -0588235294117647 16 = 9411764705882352 - - -Note that for cyclic numbers, leading zeros are important. - - -There is only one cyclic number for which, the eleven leftmost digits are 00000000137 and the five rightmost digits are 56789 (i.e., it has the form 00000000137...56789 with an unknown number of digits in the middle). Find the sum of all its digits. +A cyclic number with n digits has a very interesting property: +When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly the same digits, in the same order, but rotated in a circular fashion! + + +The smallest cyclic number is the 6-digit number 142857 : +142857 1 = 142857 +142857 2 = 285714 +142857 3 = 428571 +142857 4 = 571428 +142857 5 = 714285 +142857 6 = 857142 + + +The next cyclic number is 0588235294117647 with 16 digits : +0588235294117647 1 = 0588235294117647 +0588235294117647 2 = 1176470588235294 +0588235294117647 3 = 1764705882352941 +... +0588235294117647 16 = 9411764705882352 + + +Note that for cyclic numbers, leading zeros are important. + + +There is only one cyclic number for which, the eleven leftmost digits are 00000000137 and the five rightmost digits are 56789 (i.e., it has the form 00000000137...56789 with an unknown number of digits in the middle). Find the sum of all its digits. diff --git a/project_euler/problems/359_hilberts_new_hotel.txt b/project_euler/problems/359_hilberts_new_hotel.txt index db7a922..6082a1f 100644 --- a/project_euler/problems/359_hilberts_new_hotel.txt +++ b/project_euler/problems/359_hilberts_new_hotel.txt @@ -2,33 +2,33 @@ http://projecteuler.net/problem=359 Hilbert's New Hotel - -An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get a room at Hilbert's newest infinite hotel. The hotel contains an infinite number of floors (numbered 1, 2, 3, etc.), and each floor contains an infinite number of rooms (numbered 1, 2, 3, etc.). - -Initially the hotel is empty. Hilbert declares a rule on how the nth person is assigned a room: person n gets the first vacant room in the lowest numbered floor satisfying either of the following: +An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get a room at Hilbert's newest infinite hotel. The hotel contains an infinite number of floors (numbered 1, 2, 3, etc.), and each floor contains an infinite number of rooms (numbered 1, 2, 3, etc.). + + +Initially the hotel is empty. Hilbert declares a rule on how the nth person is assigned a room: person n gets the first vacant room in the lowest numbered floor satisfying either of the following: the floor is empty the floor is not empty, and if the latest person taking a room in that floor is person m, then m + n is a perfect square - -Person 1 gets room 1 in floor 1 since floor 1 is empty. -Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect square. -Person 2 instead gets room 1 in floor 2 since floor 2 is empty. -Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square. - - -Eventually, every person in the line gets a room in the hotel. - - -Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no person occupies the room. Here are a few examples: -P(1, 1) = 1 -P(1, 2) = 3 -P(2, 1) = 2 -P(10, 20) = 440 -P(25, 75) = 4863 -P(99, 100) = 19454 - - -Find the sum of all P(f, r) for all positive f and r such that f r = 71328803586048 and give the last 8 digits as your answer. + +Person 1 gets room 1 in floor 1 since floor 1 is empty. +Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect square. +Person 2 instead gets room 1 in floor 2 since floor 2 is empty. +Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square. + + +Eventually, every person in the line gets a room in the hotel. + + +Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no person occupies the room. Here are a few examples: +P(1, 1) = 1 +P(1, 2) = 3 +P(2, 1) = 2 +P(10, 20) = 440 +P(25, 75) = 4863 +P(99, 100) = 19454 + + +Find the sum of all P(f, r) for all positive f and r such that f r = 71328803586048 and give the last 8 digits as your answer. diff --git a/project_euler/problems/360_scary_sphere.txt b/project_euler/problems/360_scary_sphere.txt index 6a45e34..0b6ea95 100644 --- a/project_euler/problems/360_scary_sphere.txt +++ b/project_euler/problems/360_scary_sphere.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=360 Scary Sphere - -Given two points (x1,y1,z1) and (x2,y2,z2) in three dimensional space, the Manhattan distance between those points is defined as |x1-x2|+|y1-y2|+|z1-z2|. - -Let C(r) be a sphere with radius r and center in the origin O(0,0,0). -Let I(r) be the set of all points with integer coordinates on the surface of C(r). -Let S(r) be the sum of the Manhattan distances of all elements of I(r) to the origin O. +Given two points (x1,y1,z1) and (x2,y2,z2) in three dimensional space, the Manhattan distance between those points is defined as |x1-x2|+|y1-y2|+|z1-z2|. - -E.g. S(45)=34518. - -Find S(1010). +Let C(r) be a sphere with radius r and center in the origin O(0,0,0). +Let I(r) be the set of all points with integer coordinates on the surface of C(r). +Let S(r) be the sum of the Manhattan distances of all elements of I(r) to the origin O. + + +E.g. S(45)=34518. + + +Find S(1010). diff --git a/project_euler/problems/361_subsequence_of_thuemorse_sequence.txt b/project_euler/problems/361_subsequence_of_thuemorse_sequence.txt index c2b199b..ba1ef58 100644 --- a/project_euler/problems/361_subsequence_of_thuemorse_sequence.txt +++ b/project_euler/problems/361_subsequence_of_thuemorse_sequence.txt @@ -8,25 +8,25 @@ T0 = 0 T2n = Tn T2n+1 = 1 - Tn - -The first several terms of {Tn} are given as follows: -01101001100101101001011001101001.... - -We define {An} as the sorted sequence of integers such that the binary expression of each element appears as a subsequence in {Tn}. -For example, the decimal number 18 is expressed as 10010 in binary. 10010 appears in {Tn} (T8 to T12), so 18 is an element of {An}. -The decimal number 14 is expressed as 1110 in binary. 1110 never appears in {Tn}, so 14 is not an element of {An}. +The first several terms of {Tn} are given as follows: +01101001100101101001011001101001.... + + +We define {An} as the sorted sequence of integers such that the binary expression of each element appears as a subsequence in {Tn}. +For example, the decimal number 18 is expressed as 10010 in binary. 10010 appears in {Tn} (T8 to T12), so 18 is an element of {An}. +The decimal number 14 is expressed as 1110 in binary. 1110 never appears in {Tn}, so 14 is not an element of {An}. + - The first several terms of An are given as follows: n0123456789101112… An012345691011121318… - -We can also verify that A100 = 3251 and A1000 = 80852364498. - -Find the last 9 digits of . +We can also verify that A100 = 3251 and A1000 = 80852364498. + + +Find the last 9 digits of . diff --git a/project_euler/problems/362_squarefree_factors.txt b/project_euler/problems/362_squarefree_factors.txt index 7949f38..59bf4c7 100644 --- a/project_euler/problems/362_squarefree_factors.txt +++ b/project_euler/problems/362_squarefree_factors.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=362 Squarefree factors - -Consider the number 54. -54 can be factored in 7 distinct ways into one or more factors larger than 1: -54, 227, 318, 69, 336, 239 and 2333. -If we require that the factors are all squarefree only two ways remain: 336 and 2333. - -Let's call Fsf(n) the number of ways n can be factored into one or more squarefree factors larger than 1, so -Fsf(54)=2. +Consider the number 54. +54 can be factored in 7 distinct ways into one or more factors larger than 1: +54, 227, 318, 69, 336, 239 and 2333. +If we require that the factors are all squarefree only two ways remain: 336 and 2333. - -Let S(n) be Fsf(k) for k=2 to n. - -S(100)=193. +Let's call Fsf(n) the number of ways n can be factored into one or more squarefree factors larger than 1, so +Fsf(54)=2. - -Find S(10 000 000 000). + +Let S(n) be Fsf(k) for k=2 to n. + + +S(100)=193. + + +Find S(10 000 000 000). diff --git a/project_euler/problems/363_bzier_curves.txt b/project_euler/problems/363_bzier_curves.txt index 1ddfa34..617aa58 100644 --- a/project_euler/problems/363_bzier_curves.txt +++ b/project_euler/problems/363_bzier_curves.txt @@ -1,44 +1,44 @@ http://projecteuler.net/problem=363 Bézier Curves - -A cubic Bézier curve is defined by four points: P0, P1, P2 and P3. - - -The curve is constructed as follows: -On the segments P0P1, P1P2 and P2P3 the points Q0,Q1 and Q2 are drawn such that P0Q0/P0P1=P1Q1/P1P2=P2Q2/P2P3=t (t in [0,1]). -On the segments Q0Q1 and Q1Q2 the points R0 and R1 are drawn such that -Q0R0/Q0Q1=Q1R1/Q1Q2=t for the same value of t. -On the segment R0R1 the point B is drawn such that R0B/R0R1=t for the same value of t. -The Bézier curve defined by the points P0, P1, P2, P3 is the locus of B as Q0 takes all possible positions on the segment P0P1. (Please note that for all points the value of t is the same.) +A cubic Bézier curve is defined by four points: P0, P1, P2 and P3. +The curve is constructed as follows: +On the segments P0P1, P1P2 and P2P3 the points Q0,Q1 and Q2 are drawn such that P0Q0/P0P1=P1Q1/P1P2=P2Q2/P2P3=t (t in [0,1]). +On the segments Q0Q1 and Q1Q2 the points R0 and R1 are drawn such that +Q0R0/Q0Q1=Q1R1/Q1Q2=t for the same value of t. +On the segment R0R1 the point B is drawn such that R0B/R0R1=t for the same value of t. +The Bézier curve defined by the points P0, P1, P2, P3 is the locus of B as Q0 takes all possible positions on the segment P0P1. (Please note that for all points the value of t is the same.) - -In the applet to the right you can drag the points P0, P1, P2 and P3 to see what the Bézier curve (green curve) defined by those points looks like. You can also drag the point Q0 along the segment P0P1. - -From the construction it is clear that the Bézier curve will be tangent to the segments P0P1 in P0 and P2P3 in P3. +In the applet to the right you can drag the points P0, P1, P2 and P3 to see what the Bézier curve (green curve) defined by those points looks like. You can also drag the point Q0 along the segment P0P1. +From the construction it is clear that the Bézier curve will be tangent to the segments P0P1 in P0 and P2P3 in P3. + + + + + + + +A cubic Bézier curve with P0=(1,0), P1=(1,v), P2=(v,1) and P3=(0,1) is used to approximate a quarter circle. +The value v0 is chosen such that the area enclosed by the lines OP0, OP3 and the curve is equal to π/4 (the area of the quarter circle). + + +By how many percent does the length of the curve differ from the length of the quarter circle? +That is, if L is the length of the curve, calculate 100*(L-π/2)/(π/2). +Give your answer rounded to 10 digits behind the decimal point. + - -A cubic Bézier curve with P0=(1,0), P1=(1,v), P2=(v,1) and P3=(0,1) is used to approximate a quarter circle. -The value v0 is chosen such that the area enclosed by the lines OP0, OP3 and the curve is equal to π/4 (the area of the quarter circle). - -By how many percent does the length of the curve differ from the length of the quarter circle? -That is, if L is the length of the curve, calculate 100*(L-π/2)/(π/2). -Give your answer rounded to 10 digits behind the decimal point. - - - diff --git a/project_euler/problems/364_comfortable_distance.txt b/project_euler/problems/364_comfortable_distance.txt index db7e066..4492b08 100644 --- a/project_euler/problems/364_comfortable_distance.txt +++ b/project_euler/problems/364_comfortable_distance.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=364 Comfortable distance - -There are N seats in a row. N people come after each other to fill the seats according to the following rules: + +There are N seats in a row. N people come after each other to fill the seats according to the following rules: If there is any seat whose adjacent seat(s) are not occupied take such a seat. If there is no such seat and there is any seat for which only one adjacent seat is occupied take such a seat. Otherwise take one of the remaining available seats. - -Let T(N) be the number of possibilities that N seats are occupied by N people with the given rules. The following figure shows T(4)=8. - - + +Let T(N) be the number of possibilities that N seats are occupied by N people with the given rules. The following figure shows T(4)=8. + + diff --git a/project_euler/problems/365_a_huge_binomial_coefficient.txt b/project_euler/problems/365_a_huge_binomial_coefficient.txt index 266f97a..fac3c28 100644 --- a/project_euler/problems/365_a_huge_binomial_coefficient.txt +++ b/project_euler/problems/365_a_huge_binomial_coefficient.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=365 A huge binomial coefficient - -The binomial coeffient C(1018,109) is a number with more than 9 billion (9109) digits. - -Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m. +The binomial coeffient C(1018,109) is a number with more than 9 billion (9109) digits. - -Calculate M(1018,109,p*q*r) for 1000pqr5000 and p,q,r prime. + +Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m. + + +Calculate M(1018,109,p*q*r) for 1000pqr5000 and p,q,r prime. diff --git a/project_euler/problems/366_stone_game_iii.txt b/project_euler/problems/366_stone_game_iii.txt index d67de23..406687a 100644 --- a/project_euler/problems/366_stone_game_iii.txt +++ b/project_euler/problems/366_stone_game_iii.txt @@ -2,30 +2,30 @@ http://projecteuler.net/problem=366 Stone Game III - -Two players, Anton and Bernhard, are playing the following game. -There is one pile of n stones. -The first player may remove any positive number of stones, but not the whole pile. -Thereafter, each player may remove at most twice the number of stones his opponent took on the previous move. -The player who removes the last stone wins. - - -E.g. n=5 -If the first player takes anything more than one stone the next player will be able to take all remaining stones. -If the first player takes one stone, leaving four, his opponent will take also one stone, leaving three stones. -The first player cannot take all three because he may take at most 2x1=2 stones. So let's say he takes also one stone, leaving 2. The second player can take the two remaining stones and wins. -So 5 is a losing position for the first player. -For some winning positions there is more than one possible move for the first player. -E.g. when n=17 the first player can remove one or four stones. - - -Let M(n) be the maximum number of stones the first player can take from a winning position at his first turn and M(n)=0 for any other position. - - -M(n) for n100 is 728. - - -Find M(n) for n1018. -Give your answer modulo 108. + +Two players, Anton and Bernhard, are playing the following game. +There is one pile of n stones. +The first player may remove any positive number of stones, but not the whole pile. +Thereafter, each player may remove at most twice the number of stones his opponent took on the previous move. +The player who removes the last stone wins. + + +E.g. n=5 +If the first player takes anything more than one stone the next player will be able to take all remaining stones. +If the first player takes one stone, leaving four, his opponent will take also one stone, leaving three stones. +The first player cannot take all three because he may take at most 2x1=2 stones. So let's say he takes also one stone, leaving 2. The second player can take the two remaining stones and wins. +So 5 is a losing position for the first player. +For some winning positions there is more than one possible move for the first player. +E.g. when n=17 the first player can remove one or four stones. + + +Let M(n) be the maximum number of stones the first player can take from a winning position at his first turn and M(n)=0 for any other position. + + +M(n) for n100 is 728. + + +Find M(n) for n1018. +Give your answer modulo 108. diff --git a/project_euler/problems/367_bozo_sort.txt b/project_euler/problems/367_bozo_sort.txt index b4d0110..079fe5e 100644 --- a/project_euler/problems/367_bozo_sort.txt +++ b/project_euler/problems/367_bozo_sort.txt @@ -3,22 +3,22 @@ http://projecteuler.net/problem=367 Bozo sort -Bozo sort, not to be confused with the slightly less efficient bogo sort, consists out of checking if the input sequence is sorted and if not swapping randomly two elements. This is repeated until eventually the sequence is sorted. - - -If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of swaps, averaged over all 4! input sequences is 24.75. -The already sorted sequence takes 0 steps. - - -In this problem we consider the following variant on bozo sort. -If the sequence is not in order we pick three elements at random and shuffle these three elements randomly. -All 3!=6 permutations of those three elements are equally likely. -The already sorted sequence will take 0 steps. -If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of shuffles, averaged over all 4! input sequences is 27.5. -Consider as input sequences the permutations of the first 11 natural numbers. -Averaged over all 11! input sequences, what is the expected number of shuffles this sorting algorithm will perform? - - -Give your answer rounded to the nearest integer. +Bozo sort, not to be confused with the slightly less efficient bogo sort, consists out of checking if the input sequence is sorted and if not swapping randomly two elements. This is repeated until eventually the sequence is sorted. + + +If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of swaps, averaged over all 4! input sequences is 24.75. +The already sorted sequence takes 0 steps. + + +In this problem we consider the following variant on bozo sort. +If the sequence is not in order we pick three elements at random and shuffle these three elements randomly. +All 3!=6 permutations of those three elements are equally likely. +The already sorted sequence will take 0 steps. +If we consider all permutations of the first 4 natural numbers as input the expectation value of the number of shuffles, averaged over all 4! input sequences is 27.5. +Consider as input sequences the permutations of the first 11 natural numbers. +Averaged over all 11! input sequences, what is the expected number of shuffles this sorting algorithm will perform? + + +Give your answer rounded to the nearest integer. diff --git a/project_euler/problems/368_a_kempnerlike_series.txt b/project_euler/problems/368_a_kempnerlike_series.txt index 07c820b..17c518f 100644 --- a/project_euler/problems/368_a_kempnerlike_series.txt +++ b/project_euler/problems/368_a_kempnerlike_series.txt @@ -22,16 +22,16 @@ The harmonic series 1 4 -+ ... is well known to be divergent. ++ ... is well known to be divergent. - -If we however omit from this series every term where the denominator has a 9 in it, the series remarkably enough converges to approximately 22.9206766193. -This modified harmonic series is called the Kempner series. - -Let us now consider another modified harmonic series by omitting from the harmonic series every term where the denominator has 3 or more equal consecutive digits. -One can verify that out of the first 1200 terms of the harmonic series, only 20 terms will be omitted. -These 20 omitted terms are: +If we however omit from this series every term where the denominator has a 9 in it, the series remarkably enough converges to approximately 22.9206766193. +This modified harmonic series is called the Kempner series. + + +Let us now consider another modified harmonic series by omitting from the harmonic series every term where the denominator has 3 or more equal consecutive digits. +One can verify that out of the first 1200 terms of the harmonic series, only 20 terms will be omitted. +These 20 omitted terms are: @@ -149,7 +149,7 @@ These 20 omitted terms are: 1118 - and + and 1 1119 @@ -157,11 +157,11 @@ These 20 omitted terms are: . - -This series converges as well. - -Find the value the series converges to. -Give your answer rounded to 10 digits behind the decimal point. +This series converges as well. + + +Find the value the series converges to. +Give your answer rounded to 10 digits behind the decimal point. diff --git a/project_euler/problems/369_badugi.txt b/project_euler/problems/369_badugi.txt index bb7c7f5..16ac320 100644 --- a/project_euler/problems/369_badugi.txt +++ b/project_euler/problems/369_badugi.txt @@ -1,6 +1,6 @@ http://projecteuler.net/problem=369 -Badugi +Badugi In a standard 52 card deck of playing cards, a set of 4 cards is a Badugi if it contains 4 cards with no pairs and no two cards of the same suit. diff --git a/project_euler/problems/370_geometric_triangles.txt b/project_euler/problems/370_geometric_triangles.txt index 0947ee6..0f2e6e9 100644 --- a/project_euler/problems/370_geometric_triangles.txt +++ b/project_euler/problems/370_geometric_triangles.txt @@ -1,6 +1,6 @@ http://projecteuler.net/problem=370 -Geometric triangles +Geometric triangles Let us define a geometric triangle as an integer sided triangle with sides a b c so that its sides form a geometric progression, i.e. b2 = a · c .  diff --git a/project_euler/problems/371_licence_plates.txt b/project_euler/problems/371_licence_plates.txt index 1b79834..e2cacd6 100644 --- a/project_euler/problems/371_licence_plates.txt +++ b/project_euler/problems/371_licence_plates.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=371 Licence plates - -Oregon licence plates consist of three letters followed by a three digit number (each digit can be from [0..9]). -While driving to work Seth plays the following game: -Whenever the numbers of two licence plates seen on his trip add to 1000 that's a win. - -E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as he sees them in the same trip). +Oregon licence plates consist of three letters followed by a three digit number (each digit can be from [0..9]). +While driving to work Seth plays the following game: +Whenever the numbers of two licence plates seen on his trip add to 1000 that's a win. - -Find the expected number of plates he needs to see for a win. -Give your answer rounded to 8 decimal places behind the decimal point. +E.g. MIC-012 and HAN-988 is a win and RYU-500 and SET-500 too. (as long as he sees them in the same trip). -Note: We assume that each licence plate seen is equally likely to have any three digit number on it. + +Find the expected number of plates he needs to see for a win. +Give your answer rounded to 8 decimal places behind the decimal point. + + +Note: We assume that each licence plate seen is equally likely to have any three digit number on it. diff --git a/project_euler/problems/372_pencils_of_rays.txt b/project_euler/problems/372_pencils_of_rays.txt index ca2e7bb..4595654 100644 --- a/project_euler/problems/372_pencils_of_rays.txt +++ b/project_euler/problems/372_pencils_of_rays.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=372 Pencils of rays - -Let R(M, N) be the number of lattice points (x, y) which satisfy MxN, MyN and is odd. -We can verify that R(0, 100) = 3019 and R(100, 10000) = 29750422. -Find R(2·106, 109). + +Let R(M, N) be the number of lattice points (x, y) which satisfy MxN, MyN and is odd. +We can verify that R(0, 100) = 3019 and R(100, 10000) = 29750422. +Find R(2·106, 109). Note: represents the floor function. diff --git a/project_euler/problems/373_circumscribed_circles.txt b/project_euler/problems/373_circumscribed_circles.txt index 7ea0bd0..89db6af 100644 --- a/project_euler/problems/373_circumscribed_circles.txt +++ b/project_euler/problems/373_circumscribed_circles.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=373 Circumscribed Circles - -Every triangle has a circumscribed circle that goes through the three vertices. -Consider all integer sided triangles for which the radius of the circumscribed circle is integral as well. - -Let S(n) be the sum of the radii of the circumscribed circles of all such triangles for which the radius does not exceed n. +Every triangle has a circumscribed circle that goes through the three vertices. +Consider all integer sided triangles for which the radius of the circumscribed circle is integral as well. -S(100)=4950 and S(1200)=1653605. - -Find S(107). +Let S(n) be the sum of the radii of the circumscribed circles of all such triangles for which the radius does not exceed n. + +S(100)=4950 and S(1200)=1653605. + + +Find S(107). diff --git a/project_euler/problems/374_maximum_integer_partition_product.txt b/project_euler/problems/374_maximum_integer_partition_product.txt index f3ee6c7..72772c9 100644 --- a/project_euler/problems/374_maximum_integer_partition_product.txt +++ b/project_euler/problems/374_maximum_integer_partition_product.txt @@ -1,19 +1,19 @@ http://projecteuler.net/problem=374 -Maximum Integer Partition Product +Maximum Integer Partition Product An integer partition of a number n is a way of writing n as a sum of positive integers. -Partitions that differ only in the order of their summands are considered the same. +Partitions that differ only in the order of their summands are considered the same. A partition of n into distinct parts is a partition of n in which every part occurs at most once. -The partitions of 5 into distinct parts are: +The partitions of 5 into distinct parts are: 5, 4+1 and 3+2. Let f(n) be the maximum product of the parts of any such partition of n into distinct parts and let m(n) be the number of elements of any such partition of n with that product. So f(5)=6 and m(5)=2. -For n=10 the partition with the largest product is 10=2+3+5, which gives f(10)=30 and m(10)=3. +For n=10 the partition with the largest product is 10=2+3+5, which gives f(10)=30 and m(10)=3. And their product, f(10)·m(10) = 30·3 = 90 -It can be verified that +It can be verified that f(n)·m(n) for 1 n 100 = 1683550844462. -Find f(n)·m(n) for 1 n 1014. +Find f(n)·m(n) for 1 n 1014. Give your answer modulo 982451653, the 50 millionth prime. diff --git a/project_euler/problems/375_minimum_of_subsequences.txt b/project_euler/problems/375_minimum_of_subsequences.txt index e268610..d13ff13 100644 --- a/project_euler/problems/375_minimum_of_subsequences.txt +++ b/project_euler/problems/375_minimum_of_subsequences.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=375 Minimum of subsequences - -table.p375 td { - padding: 0px 3px 0px 3px; - vertical-align: bottom; - text-align: left; -} + +table.p375 td { + padding: 0px 3px 0px 3px; + vertical-align: bottom; + text-align: left; +} Let Sn be an integer sequence produced with the following pseudo-random number generator: @@ -21,11 +21,11 @@ Sn+1 Sn2 mod 50515093 - -Let A(i, j) be the minimum of the numbers Si, Si+1, ... , Sj for i j. -Let M(N) = ΣA(i, j) for 1 i j N. + +Let A(i, j) be the minimum of the numbers Si, Si+1, ... , Sj for i j. +Let M(N) = ΣA(i, j) for 1 i j N. We can verify that M(10) = 432256955 and M(10 000) = 3264567774119. - -Find M(2 000 000 000). + +Find M(2 000 000 000). diff --git a/project_euler/problems/376_nontransitive_sets_of_dice.txt b/project_euler/problems/376_nontransitive_sets_of_dice.txt index 19020fd..6465ace 100644 --- a/project_euler/problems/376_nontransitive_sets_of_dice.txt +++ b/project_euler/problems/376_nontransitive_sets_of_dice.txt @@ -2,42 +2,42 @@ http://projecteuler.net/problem=376 Nontransitive sets of dice - -Consider the following set of dice with nonstandard pips: - -Die A: 1 4 4 4 4 4 -Die B: 2 2 2 5 5 5 +Consider the following set of dice with nonstandard pips: + + +Die A: 1 4 4 4 4 4 +Die B: 2 2 2 5 5 5 Die C: 3 3 3 3 3 6 - -A game is played by two players picking a die in turn and rolling it. The player who rolls the highest value wins. - -If the first player picks die A and the second player picks die B we get +A game is played by two players picking a die in turn and rolling it. The player who rolls the highest value wins. + + +If the first player picks die A and the second player picks die B we get P(second player wins) = 7/12 > 1/2 - -If the first player picks die B and the second player picks die C we get + +If the first player picks die B and the second player picks die C we get P(second player wins) = 7/12 > 1/2 - -If the first player picks die C and the second player picks die A we get + +If the first player picks die C and the second player picks die A we get P(second player wins) = 25/36 > 1/2 - -So whatever die the first player picks, the second player can pick another die and have a larger than 50% chance of winning. -A set of dice having this property is called a nontransitive set of dice. - +So whatever die the first player picks, the second player can pick another die and have a larger than 50% chance of winning. +A set of dice having this property is called a nontransitive set of dice. + + We wish to investigate how many sets of nontransitive dice exist. We will assume the following conditions: There are three six-sided dice with each side having between 1 and N pips, inclusive. Dice with the same set of pips are equal, regardless of which side on the die the pips are located. The same pip value may appear on multiple dice; if both players roll the same value neither player wins. The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set. - -For N = 7 we find there are 9780 such sets. -How many are there for N = 30 ? + +For N = 7 we find there are 9780 such sets. +How many are there for N = 30 ? diff --git a/project_euler/problems/377_sum_of_digits_experience_13.txt b/project_euler/problems/377_sum_of_digits_experience_13.txt index f992d57..01aff3c 100644 --- a/project_euler/problems/377_sum_of_digits_experience_13.txt +++ b/project_euler/problems/377_sum_of_digits_experience_13.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=377 Sum of digits, experience 13 - -There are 16 positive integers that do not have a zero in their digits and that have a digital sum equal to 5, namely: -5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111 and 11111. -Their sum is 17891. - -Let f(n) be the sum of all positive integers that do not have a zero in their digits and have a digital sum equal to n. +There are 16 positive integers that do not have a zero in their digits and that have a digital sum equal to 5, namely: +5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111 and 11111. +Their sum is 17891. - -Find . -Give the last 9 digits as your answer. + +Let f(n) be the sum of all positive integers that do not have a zero in their digits and have a digital sum equal to n. + + +Find . +Give the last 9 digits as your answer. diff --git a/project_euler/problems/378_triangle_triples.txt b/project_euler/problems/378_triangle_triples.txt index 8cab5ea..7b93bd2 100644 --- a/project_euler/problems/378_triangle_triples.txt +++ b/project_euler/problems/378_triangle_triples.txt @@ -5,8 +5,8 @@ Triangle Triples - -Let T(n) be the nth triangle number, so T(n) = + +Let T(n) be the nth triangle number, so T(n) = @@ -14,22 +14,22 @@ n (n+1) 2 - -. +. + + + + +Let dT(n) be the number of divisors of T(n). +E.g.: +T(7) = 28 and dT(7) = 6. - -Let dT(n) be the number of divisors of T(n). -E.g.: -T(7) = 28 and dT(7) = 6. +Let Tr(n) be the number of triples (i, j, k) such that 1 i n and dT(i) > dT(j) > dT(k). +Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776. - -Let Tr(n) be the number of triples (i, j, k) such that 1 i n and dT(i) > dT(j) > dT(k). -Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776. - -Find Tr(60 000 000). -Give the last 18 digits of your answer. +Find Tr(60 000 000). +Give the last 18 digits of your answer. diff --git a/project_euler/problems/379_least_common_multiple_count.txt b/project_euler/problems/379_least_common_multiple_count.txt index e2c8e54..4c4e4b3 100644 --- a/project_euler/problems/379_least_common_multiple_count.txt +++ b/project_euler/problems/379_least_common_multiple_count.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=379 Least common multiple count - -Let f(n) be the number of couples (x,y) with x and y positive integers, x y and the least common multiple of x and y equal to n. - -Let g be the summatory function of f, i.e.: -g(n) = f(i) for 1 i n. +Let f(n) be the number of couples (x,y) with x and y positive integers, x y and the least common multiple of x and y equal to n. - -You are given that g(106) = 37429395. - -Find g(1012). +Let g be the summatory function of f, i.e.: +g(n) = f(i) for 1 i n. + + +You are given that g(106) = 37429395. + + +Find g(1012). diff --git a/project_euler/problems/380_amazing_mazes.txt b/project_euler/problems/380_amazing_mazes.txt index 0599f1a..aa0ef66 100644 --- a/project_euler/problems/380_amazing_mazes.txt +++ b/project_euler/problems/380_amazing_mazes.txt @@ -2,22 +2,22 @@ http://projecteuler.net/problem=380 Amazing Mazes! - -An mn maze is an mn rectangular grid with walls placed between grid cells such that there is exactly one path from the top-left square to any other square. The following are examples of a 912 maze and a 1520 maze: +An mn maze is an mn rectangular grid with walls placed between grid cells such that there is exactly one path from the top-left square to any other square. The following are examples of a 912 maze and a 1520 maze: - -Let C(m,n) be the number of distinct mn mazes. Mazes which can be formed by rotation and reflection from another maze are considered distinct. - -It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12) = 2.5720e46 (in scientific notation rounded to 5 significant digits). -Find C(100,500) and write your answer in scientific notation rounded to 5 significant digits. - -When giving your answer, use a lowercase e to separate mantissa and exponent. -E.g. if the answer is 1234567891011 then the answer format would be 1.2346e12. - +Let C(m,n) be the number of distinct mn mazes. Mazes which can be formed by rotation and reflection from another maze are considered distinct. + + +It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12) = 2.5720e46 (in scientific notation rounded to 5 significant digits). +Find C(100,500) and write your answer in scientific notation rounded to 5 significant digits. + + +When giving your answer, use a lowercase e to separate mantissa and exponent. +E.g. if the answer is 1234567891011 then the answer format would be 1.2346e12. + diff --git a/project_euler/problems/381_primek_factorial.txt b/project_euler/problems/381_primek_factorial.txt index 6b503ed..9ac50cf 100644 --- a/project_euler/problems/381_primek_factorial.txt +++ b/project_euler/problems/381_primek_factorial.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=381 (prime-k) factorial - -For a prime p let S(p) = ((p-k)!) mod(p) for 1 k 5. - -For example, if p=7, -(7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872. -As 872 mod(7) = 4, S(7) = 4. +For a prime p let S(p) = ((p-k)!) mod(p) for 1 k 5. - -It can be verified that S(p) = 480 for 5 p 100. - -Find S(p) for 5 p 108. +For example, if p=7, +(7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = 720+120+24+6+2 = 872. +As 872 mod(7) = 4, S(7) = 4. + + +It can be verified that S(p) = 480 for 5 p 100. + + +Find S(p) for 5 p 108. diff --git a/project_euler/problems/382_generating_polygons.txt b/project_euler/problems/382_generating_polygons.txt index 34221a2..e149304 100644 --- a/project_euler/problems/382_generating_polygons.txt +++ b/project_euler/problems/382_generating_polygons.txt @@ -2,32 +2,32 @@ http://projecteuler.net/problem=382 Generating polygons - -A polygon is a flat shape consisting of straight line segments that are joined to form a closed chain or circuit. A polygon consists of at least three sides and does not self-intersect. - +A polygon is a flat shape consisting of straight line segments that are joined to form a closed chain or circuit. A polygon consists of at least three sides and does not self-intersect. + + A set S of positive numbers is said to generate a polygon P if: - no two sides of P are the same length, - the length of every side of P is in S, and - S contains no other value. - - -For example: -The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle). -The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a quadrilateral). + no two sides of P are the same length, + the length of every side of P is in S, and + S contains no other value. + + +For example: +The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle). +The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a quadrilateral). The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all. - + Consider the sequence s, defined as follows: -s1 = 1, s2 = 2, s3 = 3 -sn = sn-1 + sn-3 for n 3. +s1 = 1, s2 = 2, s3 = 3 +sn = sn-1 + sn-3 for n 3. + + +Let Un be the set {s1, s2, ..., sn}. For example, U10 = {1, 2, 3, 4, 6, 9, 13, 19, 28, 41}. +Let f(n) be the number of subsets of Un which generate at least one polygon. +For example, f(5) = 7, f(10) = 501 and f(25) = 18635853. - -Let Un be the set {s1, s2, ..., sn}. For example, U10 = {1, 2, 3, 4, 6, 9, 13, 19, 28, 41}. -Let f(n) be the number of subsets of Un which generate at least one polygon. -For example, f(5) = 7, f(10) = 501 and f(25) = 18635853. - -Find the last 9 digits of f(1018). +Find the last 9 digits of f(1018). diff --git a/project_euler/problems/383_divisibility_comparison_between_factorials.txt b/project_euler/problems/383_divisibility_comparison_between_factorials.txt index 275f94e..85dd963 100644 --- a/project_euler/problems/383_divisibility_comparison_between_factorials.txt +++ b/project_euler/problems/383_divisibility_comparison_between_factorials.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=383 Divisibility comparison between factorials - -Let f5(n) be the largest integer x for which 5x divides n. -For example, f5(625000) = 7. - -Let T5(n) be the number of integers i which satisfy f5((2·i-1)!) 5(i!) and 1 i n. -It can be verified that T5(103) = 68 and T5(109) = 2408210. +Let f5(n) be the largest integer x for which 5x divides n. +For example, f5(625000) = 7. - -Find T5(1018). + +Let T5(n) be the number of integers i which satisfy f5((2·i-1)!) 5(i!) and 1 i n. +It can be verified that T5(103) = 68 and T5(109) = 2408210. + + +Find T5(1018). diff --git a/project_euler/problems/384_rudinshapiro_sequence.txt b/project_euler/problems/384_rudinshapiro_sequence.txt index 0631501..8b27e5a 100644 --- a/project_euler/problems/384_rudinshapiro_sequence.txt +++ b/project_euler/problems/384_rudinshapiro_sequence.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=384 Rudin-Shapiro sequence -Define the sequence a(n) as the number of adjacent pairs of ones in the binary expansion of n (possibly overlapping). +Define the sequence a(n) as the number of adjacent pairs of ones in the binary expansion of n (possibly overlapping). E.g.: a(5) = a(1012) = 0, a(6) = a(1102) = 1, a(7) = a(1112) = 2 -Define the sequence b(n) = (-1)a(n). +Define the sequence b(n) = (-1)a(n). This sequence is called the Rudin-Shapiro sequence. Also consider the summatory sequence of b(n): . -The first couple of values of these sequences are: -n        0     1     2     3     4     5     6     7 -a(n)     0     0     0     1     0     0     1     2 -b(n)     1     1     1    -1     1     1    -1     1 +The first couple of values of these sequences are: +n        0     1     2     3     4     5     6     7 +a(n)     0     0     0     1     0     0     1     2 +b(n)     1     1     1    -1     1     1    -1     1 s(n)     1     2     3     2     3     4     3     4 The sequence s(n) has the remarkable property that all elements are positive and every positive integer k occurs exactly k times. -Define g(t,c), with 1 c t, as the index in s(n) for which t occurs for the c'th time in s(n). +Define g(t,c), with 1 c t, as the index in s(n) for which t occurs for the c'th time in s(n). E.g.: g(3,3) = 6, g(4,2) = 7 and g(54321,12345) = 1220847710. -Let F(n) be the fibonacci sequence defined by: -F(0)=F(1)=1 and +Let F(n) be the fibonacci sequence defined by: +F(0)=F(1)=1 and F(n)=F(n-1)+F(n-2) for n>1. Define GF(t)=g(F(t),F(t-1)). Find ΣGF(t) for 2t45. diff --git a/project_euler/problems/385_ellipses_inside_triangles.txt b/project_euler/problems/385_ellipses_inside_triangles.txt index 41bb1db..6ddced5 100644 --- a/project_euler/problems/385_ellipses_inside_triangles.txt +++ b/project_euler/problems/385_ellipses_inside_triangles.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=385 Ellipses inside triangles - -For any triangle T in the plane, it can be shown that there is a unique ellipse with largest area that is completely inside T. +For any triangle T in the plane, it can be shown that there is a unique ellipse with largest area that is completely inside T. - -For a given n, consider triangles T such that: -- the vertices of T have integer coordinates with absolute value n, and -- the foci1 of the largest-area ellipse inside T are (13,0) and (-13,0). -Let A(n) be the sum of the areas of all such triangles. - -For example, if n = 8, there are two such triangles. Their vertices are (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each triangle is 36. Thus A(8) = 36 + 36 = 72. - -It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008. +For a given n, consider triangles T such that: +- the vertices of T have integer coordinates with absolute value n, and +- the foci1 of the largest-area ellipse inside T are (13,0) and (-13,0). +Let A(n) be the sum of the areas of all such triangles. - -Find A(1 000 000 000). + +For example, if n = 8, there are two such triangles. Their vertices are (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each triangle is 36. Thus A(8) = 36 + 36 = 72. + + +It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008. + + +Find A(1 000 000 000). + + +1The foci (plural of focus) of an ellipse are two points A and B such that for every point P on the boundary of the ellipse, AP + PB is constant. -1The foci (plural of focus) of an ellipse are two points A and B such that for every point P on the boundary of the ellipse, AP + PB is constant. - - diff --git a/project_euler/problems/386_maximum_length_of_an_antichain.txt b/project_euler/problems/386_maximum_length_of_an_antichain.txt index a8a2cf3..95a81f1 100644 --- a/project_euler/problems/386_maximum_length_of_an_antichain.txt +++ b/project_euler/problems/386_maximum_length_of_an_antichain.txt @@ -4,8 +4,8 @@ Maximum length of an antichain Let n be an integer and S(n) be the set of factors of n. A subset A of S(n) is called an antichain of S(n) if A contains only one element or if none of the elements of A divides any of the other elements of A. -For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30} -{2, 5, 6} is not an antichain of S(30). +For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30} +{2, 5, 6} is not an antichain of S(30). {2, 3, 5} is an antichain of S(30). Let N(n) be the maximum length of an antichain of S(n). Find ΣN(n) for 1 n 108 diff --git a/project_euler/problems/387_harshad_numbers.txt b/project_euler/problems/387_harshad_numbers.txt index a7e3645..44d312b 100644 --- a/project_euler/problems/387_harshad_numbers.txt +++ b/project_euler/problems/387_harshad_numbers.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=387 Harshad Numbers -A Harshad or Niven number is a number that is divisible by the sum of its digits. -201 is a Harshad number because it is divisible by 3 (the sum of its digits.) -When we truncate the last digit from 201, we get 20, which is a Harshad number. -When we truncate the last digit from 20, we get 2, which is also a Harshad number. +A Harshad or Niven number is a number that is divisible by the sum of its digits. +201 is a Harshad number because it is divisible by 3 (the sum of its digits.) +When we truncate the last digit from 201, we get 20, which is a Harshad number. +When we truncate the last digit from 20, we get 2, which is also a Harshad number. Let's call a Harshad number that, while recursively truncating the last digit, always results in a Harshad number a right truncatable Harshad number. -Also: -201/3=67 which is prime. +Also: +201/3=67 which is prime. Let's call a Harshad number that, when divided by the sum of its digits, results in a prime a strong Harshad number. -Now take the number 2011 which is prime. -When we truncate the last digit from it we get 201, a strong Harshad number that is also right truncatable. +Now take the number 2011 which is prime. +When we truncate the last digit from it we get 201, a strong Harshad number that is also right truncatable. Let's call such primes strong, right truncatable Harshad primes. You are given that the sum of the strong, right truncatable Harshad primes less than 10000 is 90619. Find the sum of the strong, right truncatable Harshad primes less than 1014. diff --git a/project_euler/problems/388_distinct_lines.txt b/project_euler/problems/388_distinct_lines.txt index 0600365..de63a78 100644 --- a/project_euler/problems/388_distinct_lines.txt +++ b/project_euler/problems/388_distinct_lines.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=388 Distinct Lines - -Consider all lattice points (a,b,c) with 0 a,b,c N. - -From the origin O(0,0,0) all lines are drawn to the other lattice points. -Let D(N) be the number of distinct such lines. +Consider all lattice points (a,b,c) with 0 a,b,c N. - -You are given that D(1 000 000) = 831909254469114121. - -Find D(1010). Give as your answer the first nine digits followed by the last nine digits. + +From the origin O(0,0,0) all lines are drawn to the other lattice points. +Let D(N) be the number of distinct such lines. + + +You are given that D(1 000 000) = 831909254469114121. + +Find D(1010). Give as your answer the first nine digits followed by the last nine digits. diff --git a/project_euler/problems/389_platonic_dice.txt b/project_euler/problems/389_platonic_dice.txt index 5e429be..09ead4e 100644 --- a/project_euler/problems/389_platonic_dice.txt +++ b/project_euler/problems/389_platonic_dice.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=389 Platonic Dice - + An unbiased single 4-sided die is thrown and its value, T, is noted. T unbiased 6-sided dice are thrown and their scores are added together. The sum, C, is noted. C unbiased 8-sided dice are thrown and their scores are added together. The sum, O, is noted. O unbiased 12-sided dice are thrown and their scores are added together. The sum, D, is noted. -D unbiased 20-sided dice are thrown and their scores are added together. The sum, I, is noted. -Find the variance of I, and give your answer rounded to 4 decimal places. +D unbiased 20-sided dice are thrown and their scores are added together. The sum, I, is noted. +Find the variance of I, and give your answer rounded to 4 decimal places. diff --git a/project_euler/problems/390_triangles_with_non_rational_sides_and_integral_area.txt b/project_euler/problems/390_triangles_with_non_rational_sides_and_integral_area.txt index 1a08d4a..8df03bd 100644 --- a/project_euler/problems/390_triangles_with_non_rational_sides_and_integral_area.txt +++ b/project_euler/problems/390_triangles_with_non_rational_sides_and_integral_area.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=390 Triangles with non rational sides and integral area -Consider the triangle with sides 5, 65 and 68. +Consider the triangle with sides 5, 65 and 68. It can be shown that this triangle has area 9. S(n) is the sum of the areas of all triangles with sides (1+b2), (1+c2) and (b2+c2) (for positive integers b and c ) that have an integral area not exceeding n. The example triangle has b=2 and c=8. diff --git a/project_euler/problems/391_hopping_game.txt b/project_euler/problems/391_hopping_game.txt index 6772487..aa45e43 100644 --- a/project_euler/problems/391_hopping_game.txt +++ b/project_euler/problems/391_hopping_game.txt @@ -2,30 +2,30 @@ http://projecteuler.net/problem=391 Hopping Game - -Let sk be the number of 1’s when writing the numbers from 0 to k in binary. -For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101. There are seven 1’s, so s5 = 7. + +Let sk be the number of 1’s when writing the numbers from 0 to k in binary. +For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101. There are seven 1’s, so s5 = 7. The sequence S = {sk : k 0} starts {0, 1, 2, 4, 5, 7, 9, 12, ...}. - -A game is played by two players. Before the game starts, a number n is chosen. A counter c starts at 0. At each turn, the player chooses a number from 1 to n (inclusive) and increases c by that number. The resulting value of c must be a member of S. If there are no more valid moves, the player loses. - -For example: -Let n = 5. c starts at 0. -Player 1 chooses 4, so c becomes 0 + 4 = 4. -Player 2 chooses 5, so c becomes 4 + 5 = 9. -Player 1 chooses 3, so c becomes 9 + 3 = 12. -etc. -Note that c must always belong to S, and each player can increase c by at most n. +A game is played by two players. Before the game starts, a number n is chosen. A counter c starts at 0. At each turn, the player chooses a number from 1 to n (inclusive) and increases c by that number. The resulting value of c must be a member of S. If there are no more valid moves, the player loses. + + +For example: +Let n = 5. c starts at 0. +Player 1 chooses 4, so c becomes 0 + 4 = 4. +Player 2 chooses 5, so c becomes 4 + 5 = 9. +Player 1 chooses 3, so c becomes 9 + 3 = 12. +etc. +Note that c must always belong to S, and each player can increase c by at most n. + + +Let M(n) be the highest number the first player can choose at her first turn to force a win, and M(n) = 0 if there is no such move. For example, M(2) = 2, M(7) = 1 and M(20) = 4. + - -Let M(n) be the highest number the first player can choose at her first turn to force a win, and M(n) = 0 if there is no such move. For example, M(2) = 2, M(7) = 1 and M(20) = 4. +Given Σ(M(n))3 = 8150 for 1 n 20. - -Given Σ(M(n))3 = 8150 for 1 n 20. - -Find Σ(M(n))3 for 1 n 1000. +Find Σ(M(n))3 for 1 n 1000. diff --git a/project_euler/problems/392_enmeshed_unit_circle.txt b/project_euler/problems/392_enmeshed_unit_circle.txt index 7c55726..dda5c39 100644 --- a/project_euler/problems/392_enmeshed_unit_circle.txt +++ b/project_euler/problems/392_enmeshed_unit_circle.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=392 Enmeshed unit circle - -A rectilinear grid is an orthogonal grid where the spacing between the gridlines does not have to be equidistant. -An example of such grid is logarithmic graph paper. - - -Consider rectilinear grids in the Cartesian coordinate system with the following properties:The gridlines are parallel to the axes of the Cartesian coordinate system.There are N+2 vertical and N+2 horizontal gridlines. Hence there are (N+1) x (N+1) rectangular cells.The equations of the two outer vertical gridlines are x = -1 and x = 1.The equations of the two outer horizontal gridlines are y = -1 and y = 1.The grid cells are colored red if they overlap with the unit circle, black otherwise.For this problem we would like you to find the postions of the remaining N inner horizontal and N inner vertical gridlines so that the area occupied by the red cells is minimized. - - -E.g. here is a picture of the solution for N = 10: - - - - -The area occupied by the red cells for N = 10 rounded to 10 digits behind the decimal point is 3.3469640797. - - -Find the positions for N = 400. -Give as your answer the area occupied by the red cells rounded to 10 digits behind the decimal point. + +A rectilinear grid is an orthogonal grid where the spacing between the gridlines does not have to be equidistant. +An example of such grid is logarithmic graph paper. + + +Consider rectilinear grids in the Cartesian coordinate system with the following properties:The gridlines are parallel to the axes of the Cartesian coordinate system.There are N+2 vertical and N+2 horizontal gridlines. Hence there are (N+1) x (N+1) rectangular cells.The equations of the two outer vertical gridlines are x = -1 and x = 1.The equations of the two outer horizontal gridlines are y = -1 and y = 1.The grid cells are colored red if they overlap with the unit circle, black otherwise.For this problem we would like you to find the postions of the remaining N inner horizontal and N inner vertical gridlines so that the area occupied by the red cells is minimized. + + +E.g. here is a picture of the solution for N = 10: + + + + +The area occupied by the red cells for N = 10 rounded to 10 digits behind the decimal point is 3.3469640797. + + +Find the positions for N = 400. +Give as your answer the area occupied by the red cells rounded to 10 digits behind the decimal point. diff --git a/project_euler/problems/393_migrating_ants.txt b/project_euler/problems/393_migrating_ants.txt index 938dc75..79de77c 100644 --- a/project_euler/problems/393_migrating_ants.txt +++ b/project_euler/problems/393_migrating_ants.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=393 Migrating ants - -An nn grid of squares contains n2 ants, one ant per square. -All ants decide to move simultaneously to an adjacent square (usually 4 possibilities, except for ants on the edge of the grid or at the corners). -We define f(n) to be the number of ways this can happen without any ants ending on the same square and without any two ants crossing the same edge between two squares. - - -You are given that f(4) = 88. -Find f(10). + +An nn grid of squares contains n2 ants, one ant per square. +All ants decide to move simultaneously to an adjacent square (usually 4 possibilities, except for ants on the edge of the grid or at the corners). +We define f(n) to be the number of ways this can happen without any ants ending on the same square and without any two ants crossing the same edge between two squares. + + +You are given that f(4) = 88. +Find f(10). diff --git a/project_euler/problems/394_eating_pie.txt b/project_euler/problems/394_eating_pie.txt index 6e8dc39..fe5206d 100644 --- a/project_euler/problems/394_eating_pie.txt +++ b/project_euler/problems/394_eating_pie.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=394 Eating pie - -Jeff eats a pie in an unusual way. -The pie is circular. He starts with slicing an initial cut in the pie along a radius. -While there is at least a given fraction F of pie left, he performs the following procedure: -- He makes two slices from the pie centre to any point of what is remaining of the pie border, any point on the remaining pie border equally likely. This will divide the remaining pie into three pieces. -- Going counterclockwise from the initial cut, he takes the first two pie pieces and eats them. -When less than a fraction F of pie remains, he does not repeat this procedure. Instead, he eats all of the remaining pie. +Jeff eats a pie in an unusual way. +The pie is circular. He starts with slicing an initial cut in the pie along a radius. +While there is at least a given fraction F of pie left, he performs the following procedure: +- He makes two slices from the pie centre to any point of what is remaining of the pie border, any point on the remaining pie border equally likely. This will divide the remaining pie into three pieces. +- Going counterclockwise from the initial cut, he takes the first two pie pieces and eats them. +When less than a fraction F of pie remains, he does not repeat this procedure. Instead, he eats all of the remaining pie. - -For x 1, let E(x) be the expected number of times Jeff repeats the procedure above with F = 1/x. -It can be verified that E(1) = 1, E(2) 1.2676536759, and E(7.5) 2.1215732071. - -Find E(40) rounded to 10 decimal places behind the decimal point. + +For x 1, let E(x) be the expected number of times Jeff repeats the procedure above with F = 1/x. +It can be verified that E(1) = 1, E(2) 1.2676536759, and E(7.5) 2.1215732071. + + +Find E(40) rounded to 10 decimal places behind the decimal point. diff --git a/project_euler/problems/395_pythagorean_tree.txt b/project_euler/problems/395_pythagorean_tree.txt index 894ec6c..13b214c 100644 --- a/project_euler/problems/395_pythagorean_tree.txt +++ b/project_euler/problems/395_pythagorean_tree.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=395 Pythagorean tree - -The Pythagorean tree is a fractal generated by the following procedure: - -Start with a unit square. Then, calling one of the sides its base (in the animation, the bottom side is the base): +The Pythagorean tree is a fractal generated by the following procedure: + + +Start with a unit square. Then, calling one of the sides its base (in the animation, the bottom side is the base): Attach a right triangle to the side opposite the base, with the hypotenuse coinciding with that side and with the sides in a 3-4-5 ratio. Note that the smaller side of the triangle must be on the 'right' side with respect to the base (see animation). Attach a square to each leg of the right triangle, with one of its sides coinciding with that leg. Repeat this procedure for both squares, considering as their bases the sides touching the triangle. - -The resulting figure, after an infinite number of iterations, is the Pythagorean tree. - - - -It can be shown that there exists at least one rectangle, whose sides are parallel to the largest square of the Pythagorean tree, which encloses the Pythagorean tree completely. +The resulting figure, after an infinite number of iterations, is the Pythagorean tree. + + + + +It can be shown that there exists at least one rectangle, whose sides are parallel to the largest square of the Pythagorean tree, which encloses the Pythagorean tree completely. + - -Find the smallest area possible for such a bounding rectangle, and give your answer rounded to 10 decimal places. +Find the smallest area possible for such a bounding rectangle, and give your answer rounded to 10 decimal places. diff --git a/project_euler/problems/396_weak_goodstein_sequence.txt b/project_euler/problems/396_weak_goodstein_sequence.txt index 4e4c89b..c60c97b 100644 --- a/project_euler/problems/396_weak_goodstein_sequence.txt +++ b/project_euler/problems/396_weak_goodstein_sequence.txt @@ -2,33 +2,33 @@ http://projecteuler.net/problem=396 Weak Goodstein sequence - -For any positive integer n, the nth weak Goodstein sequence {g1, g2, g3, ...} is defined as: + +For any positive integer n, the nth weak Goodstein sequence {g1, g2, g3, ...} is defined as: g1 = n - for k 1, gk is obtained by writing gk-1 in base k, interpreting it as a base k + 1 number, and subtracting 1. - -The sequence terminates when gk becomes 0. - - -For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}: - - g1 = 6. - g2 = 11 since 6 = 1102, 1103 = 12, and 12 - 1 = 11. - g3 = 17 since 11 = 1023, 1024 = 18, and 18 - 1 = 17. - g4 = 25 since 17 = 1014, 1015 = 26, and 26 - 1 = 25. - -and so on. - - -It can be shown that every weak Goodstein sequence terminates. - - -Let G(n) be the number of nonzero elements in the nth weak Goodstein sequence. -It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381. -It can also be verified that ΣG(n) = 2517 for 1 n 8. - - -Find the last 9 digits of ΣG(n) for 1 n 16. + for k 1, gk is obtained by writing gk-1 in base k, interpreting it as a base k + 1 number, and subtracting 1. + +The sequence terminates when gk becomes 0. + + +For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}: + + g1 = 6. + g2 = 11 since 6 = 1102, 1103 = 12, and 12 - 1 = 11. + g3 = 17 since 11 = 1023, 1024 = 18, and 18 - 1 = 17. + g4 = 25 since 17 = 1014, 1015 = 26, and 26 - 1 = 25. + +and so on. + + +It can be shown that every weak Goodstein sequence terminates. + + +Let G(n) be the number of nonzero elements in the nth weak Goodstein sequence. +It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381. +It can also be verified that ΣG(n) = 2517 for 1 n 8. + + +Find the last 9 digits of ΣG(n) for 1 n 16. diff --git a/project_euler/problems/397_triangle_on_parabola.txt b/project_euler/problems/397_triangle_on_parabola.txt index 1d1e64c..cd9d9f6 100644 --- a/project_euler/problems/397_triangle_on_parabola.txt +++ b/project_euler/problems/397_triangle_on_parabola.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=397 Triangle on parabola - -On the parabola y = x2/k, three points A(a, a2/k), B(b, b2/k) and C(c, c2/k) are chosen. - -Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such that at least one angle of the triangle ABC is 45-degree, with 1 k K and -X a b c X. +On the parabola y = x2/k, three points A(a, a2/k), B(b, b2/k) and C(c, c2/k) are chosen. - -For example, F(1, 10) = 41 and F(10, 100) = 12492. -Find F(106, 109). + +Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such that at least one angle of the triangle ABC is 45-degree, with 1 k K and -X a b c X. + + +For example, F(1, 10) = 41 and F(10, 100) = 12492. +Find F(106, 109). diff --git a/project_euler/problems/398_cutting_rope.txt b/project_euler/problems/398_cutting_rope.txt index a2f2e7f..186b9cd 100644 --- a/project_euler/problems/398_cutting_rope.txt +++ b/project_euler/problems/398_cutting_rope.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=398 Cutting rope - -Inside a rope of length n, n-1 points are placed with distance 1 from each other and from the endpoints. Among these points, we choose m-1 points at random and cut the rope at these points to create m segments. - -Let E(n, m) be the expected length of the second-shortest segment. -For example, E(3, 2) = 2 and E(8, 3) = 16/7. -Note that if multiple segments have the same shortest length the length of the second-shortest segment is defined as the same as the shortest length. +Inside a rope of length n, n-1 points are placed with distance 1 from each other and from the endpoints. Among these points, we choose m-1 points at random and cut the rope at these points to create m segments. - -Find E(107, 100). -Give your answer rounded to 5 decimal places behind the decimal point. + +Let E(n, m) be the expected length of the second-shortest segment. +For example, E(3, 2) = 2 and E(8, 3) = 16/7. +Note that if multiple segments have the same shortest length the length of the second-shortest segment is defined as the same as the shortest length. + + +Find E(107, 100). +Give your answer rounded to 5 decimal places behind the decimal point. diff --git a/project_euler/problems/399_squarefree_fibonacci_numbers.txt b/project_euler/problems/399_squarefree_fibonacci_numbers.txt index dda0a43..3d8ed51 100644 --- a/project_euler/problems/399_squarefree_fibonacci_numbers.txt +++ b/project_euler/problems/399_squarefree_fibonacci_numbers.txt @@ -2,29 +2,29 @@ http://projecteuler.net/problem=399 Squarefree Fibonacci Numbers - -The first 15 fibonacci numbers are: -1,1,2,3,5,8,13,21,34,55,89,144,233,377,610. -It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and 144 is divisible by 4 and by 9. -So the first 13 squarefree fibonacci numbers are: -1,1,2,3,5,13,21,34,55,89,233,377 and 610. - - -The 200th squarefree fibonacci number is: -971183874599339129547649988289594072811608739584170445. -The last sixteen digits of this number are: 1608739584170445 and in scientific notation this number can be written as 9.7e53. - - -Find the 100 000 000th squarefree fibonacci number. -Give as your answer its last sixteen digits followed by a comma followed by the number in scientific notation (rounded to one digit after the decimal point). -For the 200th squarefree number the answer would have been: 1608739584170445,9.7e53 - - - -Note: -For this problem, assume that for every prime p, the first fibonacci number divisible by p is not divisible by p2 (this is part of Wall's conjecture). This has been verified for primes 3·1015, but has not been proven in general. - -If it happens that the conjecture is false, then the accepted answer to this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci number, rather it represents only a lower bound for that number. + +The first 15 fibonacci numbers are: +1,1,2,3,5,8,13,21,34,55,89,144,233,377,610. +It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and 144 is divisible by 4 and by 9. +So the first 13 squarefree fibonacci numbers are: +1,1,2,3,5,13,21,34,55,89,233,377 and 610. + + +The 200th squarefree fibonacci number is: +971183874599339129547649988289594072811608739584170445. +The last sixteen digits of this number are: 1608739584170445 and in scientific notation this number can be written as 9.7e53. + + +Find the 100 000 000th squarefree fibonacci number. +Give as your answer its last sixteen digits followed by a comma followed by the number in scientific notation (rounded to one digit after the decimal point). +For the 200th squarefree number the answer would have been: 1608739584170445,9.7e53 + + + +Note: +For this problem, assume that for every prime p, the first fibonacci number divisible by p is not divisible by p2 (this is part of Wall's conjecture). This has been verified for primes 3·1015, but has not been proven in general. + +If it happens that the conjecture is false, then the accepted answer to this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci number, rather it represents only a lower bound for that number. diff --git a/project_euler/problems/400_fibonacci_tree_game.txt b/project_euler/problems/400_fibonacci_tree_game.txt index c139eb8..b0c57a7 100644 --- a/project_euler/problems/400_fibonacci_tree_game.txt +++ b/project_euler/problems/400_fibonacci_tree_game.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=400 Fibonacci tree game - + A Fibonacci tree is a binary tree recursively defined as: -T(0) is the empty tree. -T(1) is the binary tree with only one node. -T(k) consists of a root node that has T(k-1) and T(k-2) as children. +T(0) is the empty tree. +T(1) is the binary tree with only one node. +T(k) consists of a root node that has T(k-1) and T(k-2) as children. + - -On such a tree two players play a take-away game. On each turn a player selects a node and removes that node along with the subtree rooted at that node. +On such a tree two players play a take-away game. On each turn a player selects a node and removes that node along with the subtree rooted at that node. The player who is forced to take the root node of the entire tree loses. - -Here are the winning moves of the first player on the first turn for T(k) from k=1 to k=6. - - - - -Let f(k) be the number of winning moves of the first player (i.e. the moves for which the second player has no winning strategy) on the first turn of the game when this game is played on T(k). - - - -For example, f(5) = 1 and f(10) = 17. - - -Find f(10000). Give the last 18 digits of your answer. + +Here are the winning moves of the first player on the first turn for T(k) from k=1 to k=6. + + + + +Let f(k) be the number of winning moves of the first player (i.e. the moves for which the second player has no winning strategy) on the first turn of the game when this game is played on T(k). + + + +For example, f(5) = 1 and f(10) = 17. + + +Find f(10000). Give the last 18 digits of your answer. diff --git a/project_euler/problems/401_sum_of_squares_of_divisors.txt b/project_euler/problems/401_sum_of_squares_of_divisors.txt index 406521b..22746f6 100644 --- a/project_euler/problems/401_sum_of_squares_of_divisors.txt +++ b/project_euler/problems/401_sum_of_squares_of_divisors.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=401 Sum of squares of divisors - -The divisors of 6 are 1,2,3 and 6. -The sum of the squares of these numbers is 1+4+9+36=50. - - -Let sigma2(n) represent the sum of the squares of the divisors of n. -Thus sigma2(6)=50. - -Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=sigma2(i) for i=1 to n. -The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113. - - -Find SIGMA2(1015) modulo 109. + +The divisors of 6 are 1,2,3 and 6. +The sum of the squares of these numbers is 1+4+9+36=50. + + +Let sigma2(n) represent the sum of the squares of the divisors of n. +Thus sigma2(6)=50. + +Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=sigma2(i) for i=1 to n. +The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113. + + +Find SIGMA2(1015) modulo 109. diff --git a/project_euler/problems/402_integervalued_polynomials.txt b/project_euler/problems/402_integervalued_polynomials.txt index aa851b6..47f358c 100644 --- a/project_euler/problems/402_integervalued_polynomials.txt +++ b/project_euler/problems/402_integervalued_polynomials.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=402 Integer-valued polynomials - -It can be shown that the polynomial n4 + 4n3 + 2n2 + 5n is a multiple of 6 for every integer n. It can also be shown that 6 is the largest integer satisfying this property. - -Define M(a, b, c) as the maximum m such that n4 + an3 + bn2 + cn is a multiple of m for all integers n. For example, M(4, 2, 5) = 6. +It can be shown that the polynomial n4 + 4n3 + 2n2 + 5n is a multiple of 6 for every integer n. It can also be shown that 6 is the largest integer satisfying this property. - -Also, define S(N) as the sum of M(a, b, c) for all 0 a, b, c N. - -We can verify that S(10) = 1972 and S(10000) = 2024258331114. +Define M(a, b, c) as the maximum m such that n4 + an3 + bn2 + cn is a multiple of m for all integers n. For example, M(4, 2, 5) = 6. - -Let Fk be the Fibonacci sequence: -F0 = 0, F1 = 1 and -Fk = Fk-1 + Fk-2 for k 2. - -Find the last 9 digits of Σ S(Fk) for 2 k 1234567890123. +Also, define S(N) as the sum of M(a, b, c) for all 0 a, b, c N. + + +We can verify that S(10) = 1972 and S(10000) = 2024258331114. + + +Let Fk be the Fibonacci sequence: +F0 = 0, F1 = 1 and +Fk = Fk-1 + Fk-2 for k 2. + + +Find the last 9 digits of Σ S(Fk) for 2 k 1234567890123. diff --git a/project_euler/problems/403_lattice_points_enclosed_by_parabola_and_line.txt b/project_euler/problems/403_lattice_points_enclosed_by_parabola_and_line.txt index fa56c96..b3507d1 100644 --- a/project_euler/problems/403_lattice_points_enclosed_by_parabola_and_line.txt +++ b/project_euler/problems/403_lattice_points_enclosed_by_parabola_and_line.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=403 Lattice points enclosed by parabola and line - + For integers a and b, we define D(a, b) as the domain enclosed by the parabola y = x2 and the line y = a·x + b: -D(a, b) = { (x, y) | x2 y a·x + b }. +D(a, b) = { (x, y) | x2 y a·x + b }. + + +L(a, b) is defined as the number of lattice points contained in D(a, b). +For example, L(1, 2) = 8 and L(2, -1) = 1. + - -L(a, b) is defined as the number of lattice points contained in D(a, b). -For example, L(1, 2) = 8 and L(2, -1) = 1. +We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such that the area of D(a, b) is a rational number and |a|,|b| N. +We can verify that S(5) = 344 and S(100) = 26709528. - -We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such that the area of D(a, b) is a rational number and |a|,|b| N. -We can verify that S(5) = 344 and S(100) = 26709528. - -Find S(1012). Give your answer mod 108. +Find S(1012). Give your answer mod 108. diff --git a/project_euler/problems/404_crisscross_ellipses.txt b/project_euler/problems/404_crisscross_ellipses.txt index 1dc46ce..b7b3f76 100644 --- a/project_euler/problems/404_crisscross_ellipses.txt +++ b/project_euler/problems/404_crisscross_ellipses.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=404 Crisscross Ellipses - -Ea is an ellipse with an equation of the form x2 + 4y2 = 4a2. -Ea' is the rotated image of Ea by θ degrees counterclockwise around the origin O(0, 0) for 0° θ 90°. +Ea is an ellipse with an equation of the form x2 + 4y2 = 4a2. +Ea' is the rotated image of Ea by θ degrees counterclockwise around the origin O(0, 0) for 0° θ 90°. -b is the distance to the origin of the two intersection points closest to the origin and c is the distance of the two other intersection points. -We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a, b and c are positive integers. -For example, (209, 247, 286) is a canonical ellipsoidal triplet. - -Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b, c) for a N. -It can be verified that C(103) = 7, C(104) = 106 and C(106) = 11845. +b is the distance to the origin of the two intersection points closest to the origin and c is the distance of the two other intersection points. +We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a, b and c are positive integers. +For example, (209, 247, 286) is a canonical ellipsoidal triplet. - -Find C(1017). + +Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b, c) for a N. +It can be verified that C(103) = 7, C(104) = 106 and C(106) = 11845. + + +Find C(1017). diff --git a/project_euler/problems/405_a_rectangular_tiling.txt b/project_euler/problems/405_a_rectangular_tiling.txt index 909428c..a65aabf 100644 --- a/project_euler/problems/405_a_rectangular_tiling.txt +++ b/project_euler/problems/405_a_rectangular_tiling.txt @@ -2,25 +2,25 @@ http://projecteuler.net/problem=405 A rectangular tiling - -We wish to tile a rectangle whose length is twice its width. -Let T(0) be the tiling consisting of a single rectangle. -For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the following manner: +We wish to tile a rectangle whose length is twice its width. +Let T(0) be the tiling consisting of a single rectangle. +For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the following manner: - -The following animation demonstrates the tilings T(n) for n from 0 to 5: +The following animation demonstrates the tilings T(n) for n from 0 to 5: - -Let f(n) be the number of points where four tiles meet in T(n). -For example, f(1) = 0, f(4) = 82 and f(109) mod 177 = 126897180. - -Find f(10k) for k = 1018, give your answer modulo 177. + + +Let f(n) be the number of points where four tiles meet in T(n). +For example, f(1) = 0, f(4) = 82 and f(109) mod 177 = 126897180. + + +Find f(10k) for k = 1018, give your answer modulo 177. diff --git a/project_euler/problems/406_guessing_game.txt b/project_euler/problems/406_guessing_game.txt index 6ba346d..9ddb575 100644 --- a/project_euler/problems/406_guessing_game.txt +++ b/project_euler/problems/406_guessing_game.txt @@ -2,7 +2,7 @@ http://projecteuler.net/problem=406 Guessing Game -We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions. +We are trying to find a hidden number selected from the set of integers {1, 2, ..., n} by asking questions. Each number (question) we ask, we get one of three possible answers: "Your guess is lower than the hidden number" (and you incur a cost of a), or @@ -11,17 +11,17 @@ Each number (question) we ask, we get one of three possible answers: Given the value of n, a, and b, an optimal strategy minimizes the total cost for the worst possible case. For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2" as our first question. -If we are told that 2 is higher than the hidden number (for a cost of b=3), then we are sure that "1" is the hidden number (for a total cost of 3). -If we are told that 2 is lower than the hidden number (for a cost of a=2), then our next question will be "4". -If we are told that 4 is higher than the hidden number (for a cost of b=3), then we are sure that "3" is the hidden number (for a total cost of 2+3=5). -If we are told that 4 is lower than the hidden number (for a cost of a=2), then we are sure that "5" is the hidden number (for a total cost of 2+2=4). -Thus, the worst-case cost achieved by this strategy is 5. It can also be shown that this is the lowest worst-case cost that can be achieved. +If we are told that 2 is higher than the hidden number (for a cost of b=3), then we are sure that "1" is the hidden number (for a total cost of 3). +If we are told that 2 is lower than the hidden number (for a cost of a=2), then our next question will be "4". +If we are told that 4 is higher than the hidden number (for a cost of b=3), then we are sure that "3" is the hidden number (for a total cost of 2+3=5). +If we are told that 4 is lower than the hidden number (for a cost of a=2), then we are sure that "5" is the hidden number (for a total cost of 2+2=4). +Thus, the worst-case cost achieved by this strategy is 5. It can also be shown that this is the lowest worst-case cost that can be achieved. So, in fact, we have just described an optimal strategy for the given values of n, a, and b. Let C(n, a, b) be the worst-case cost achieved by an optimal strategy for the given values of n, a, and b. -Here are a few examples: -C(5, 2, 3) = 5 -C(500, 2, 3) = 13.22073197... -C(20000, 5, 7) = 82 +Here are a few examples: +C(5, 2, 3) = 5 +C(500, 2, 3) = 13.22073197... +C(20000, 5, 7) = 82 C(2000000, 5, 7) = 49.63755955... Let Fk be the Fibonacci numbers: Fk = Fk-1 + Fk-2 with base cases F1 = F2 = 1.Find 1k30 C(1012, k, Fk), and give your answer rounded to 8 decimal places behind the decimal point. diff --git a/project_euler/problems/407_idempotents.txt b/project_euler/problems/407_idempotents.txt index 1f0f3aa..9eb0323 100644 --- a/project_euler/problems/407_idempotents.txt +++ b/project_euler/problems/407_idempotents.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=407 Idempotents - -If we calculate a2 mod 6 for 0 a 5 we get: 0,1,4,3,4,1. - -The largest value of a such that a2 a mod 6 is 4. -Let's call M(n) the largest value of a n such that a2 a (mod n). -So M(6) = 4. +If we calculate a2 mod 6 for 0 a 5 we get: 0,1,4,3,4,1. - -Find M(n) for 1 n 107. + +The largest value of a such that a2 a mod 6 is 4. +Let's call M(n) the largest value of a n such that a2 a (mod n). +So M(6) = 4. + + +Find M(n) for 1 n 107. diff --git a/project_euler/problems/408_admissible_paths_through_a_grid.txt b/project_euler/problems/408_admissible_paths_through_a_grid.txt index b09fca2..224b33f 100644 --- a/project_euler/problems/408_admissible_paths_through_a_grid.txt +++ b/project_euler/problems/408_admissible_paths_through_a_grid.txt @@ -2,11 +2,11 @@ http://projecteuler.net/problem=408 Admissible paths through a grid -Let's call a lattice point (x, y) inadmissible if x, y and x + y are all positive perfect squares. +Let's call a lattice point (x, y) inadmissible if x, y and x + y are all positive perfect squares. For example, (9, 16) is inadmissible, while (0, 4), (3, 1) and (9, 4) are not. -Consider a path from point (x1, y1) to point (x2, y2) using only unit steps north or east. +Consider a path from point (x1, y1) to point (x2, y2) using only unit steps north or east. Let's call such a path admissible if none of its intermediate points are inadmissible. -Let P(n) be the number of admissible paths from (0, 0) to (n, n). +Let P(n) be the number of admissible paths from (0, 0) to (n, n). It can be verified that P(5) = 252, P(16) = 596994440 and P(1000) mod 1 000 000 007 = 341920854. Find P(10 000 000) mod 1 000 000 007. diff --git a/project_euler/problems/409_nim_extreme.txt b/project_euler/problems/409_nim_extreme.txt index eb3ee37..35e3503 100644 --- a/project_euler/problems/409_nim_extreme.txt +++ b/project_euler/problems/409_nim_extreme.txt @@ -3,13 +3,13 @@ http://projecteuler.net/problem=409 Nim Extreme Let n be a positive integer. Consider nim positions where: -There are n non-empty piles. -Each pile has size less than 2n. -No two piles have the same size. +There are n non-empty piles. +Each pile has size less than 2n. +No two piles have the same size. -Let W(n) be the number of winning nim positions satisfying the above -conditions (a position is winning if the first player has a winning strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360 and W(100) mod 1 000 000 007 = 384777056. +Let W(n) be the number of winning nim positions satisfying the above +conditions (a position is winning if the first player has a winning strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360 and W(100) mod 1 000 000 007 = 384777056. -Find W(10 000 000) mod 1 000 000 007. +Find W(10 000 000) mod 1 000 000 007. diff --git a/project_euler/problems/410_circle_and_tangent_line.txt b/project_euler/problems/410_circle_and_tangent_line.txt index 86bc56e..efaa0c2 100644 --- a/project_euler/problems/410_circle_and_tangent_line.txt +++ b/project_euler/problems/410_circle_and_tangent_line.txt @@ -5,6 +5,6 @@ Circle and tangent line Let C be the circle with radius r, x2 + y2 = r2. We choose two points P(a, b) and Q(-a, c) so that the line passing through P and Q is tangent to C. For example, the quadruplet (r, a, b, c) = (2, 6, 2, -7) satisfies this property. Let F(R, X) be the number of the integer quadruplets (r, a, b, c) with this property, and with 0 r R and 0 a X. -We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384. +We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384. Find F(108, 109) + F(109, 108). diff --git a/project_euler/problems/411_uphill_paths.txt b/project_euler/problems/411_uphill_paths.txt index 1dd9790..4976515 100644 --- a/project_euler/problems/411_uphill_paths.txt +++ b/project_euler/problems/411_uphill_paths.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=411 Uphill paths - -Let n be a positive integer. Suppose there are stations at the coordinates (x, y) = (2i mod n, 3i mod n) for 0 i 2n. We will consider stations with the same coordinates as the same station. - -We wish to form a path from (0, 0) to (n, n) such that the x and y coordinates never decrease. -Let S(n) be the maximum number of stations such a path can pass through. - -For example, if n = 22, there are 11 distinct stations, and a valid path can pass through at most 5 stations. Therefore, S(22) = 5. -The case is illustrated below, with an example of an optimal path: - - - -It can also be verified that S(123) = 14 and S(10000) = 48. - -Find S(k5) for 1 k 30. + +Let n be a positive integer. Suppose there are stations at the coordinates (x, y) = (2i mod n, 3i mod n) for 0 i 2n. We will consider stations with the same coordinates as the same station. + +We wish to form a path from (0, 0) to (n, n) such that the x and y coordinates never decrease. +Let S(n) be the maximum number of stations such a path can pass through. + +For example, if n = 22, there are 11 distinct stations, and a valid path can pass through at most 5 stations. Therefore, S(22) = 5. +The case is illustrated below, with an example of an optimal path: + + + +It can also be verified that S(123) = 14 and S(10000) = 48. + +Find S(k5) for 1 k 30. diff --git a/project_euler/problems/412_gnomon_numbering.txt b/project_euler/problems/412_gnomon_numbering.txt index e611692..0eb284b 100644 --- a/project_euler/problems/412_gnomon_numbering.txt +++ b/project_euler/problems/412_gnomon_numbering.txt @@ -8,7 +8,7 @@ For example, L(5, 3) looks like this: We want to number each cell of L(m, n) with consecutive integers 1, 2, 3, ... such that the number in every cell is smaller than the number below it and to the left of it. For example, here are two valid numberings of L(5, 3): -Let LC(m, n) be the number of valid numberings of L(m, n). +Let LC(m, n) be the number of valid numberings of L(m, n). It can be verified that LC(3, 0) = 42, LC(5, 3) = 250250, LC(6, 3) = 406029023400 and LC(10, 5) mod 76543217 = 61251715. Find LC(10000, 5000) mod 76543217. diff --git a/project_euler/problems/413_onechild_numbers.txt b/project_euler/problems/413_onechild_numbers.txt index 4802773..9ccd187 100644 --- a/project_euler/problems/413_onechild_numbers.txt +++ b/project_euler/problems/413_onechild_numbers.txt @@ -3,10 +3,10 @@ http://projecteuler.net/problem=413 One-child Numbers We say that a d-digit positive number (no leading zeros) is a one-child number if exactly one of its sub-strings is divisible by d. -For example, 5671 is a 4-digit one-child number. Among all its sub-strings 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4. -Similarly, 104 is a 3-digit one-child number because only 0 is divisible by 3. +For example, 5671 is a 4-digit one-child number. Among all its sub-strings 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4. +Similarly, 104 is a 3-digit one-child number because only 0 is divisible by 3. 1132451 is a 7-digit one-child number because only 245 is divisible by 7. -Let F(N) be the number of the one-child numbers less than N. +Let F(N) be the number of the one-child numbers less than N. We can verify that F(10) = 9, F(103) = 389 and F(107) = 277674. Find F(1019). diff --git a/project_euler/problems/414_kaprekar_constant.txt b/project_euler/problems/414_kaprekar_constant.txt index 2443949..47e3cf6 100644 --- a/project_euler/problems/414_kaprekar_constant.txt +++ b/project_euler/problems/414_kaprekar_constant.txt @@ -2,40 +2,40 @@ http://projecteuler.net/problem=414 Kaprekar constant - -6174 is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get 7641-1467=6174. -Even more remarkable is that if we start from any 4 digit number and repeat this process of sorting and subtracting, we'll eventually end up with 6174 or immediately with 0 if all digits are equal. -This also works with numbers that have less than 4 digits if we pad the number with leading zeroes until we have 4 digits. -E.g. let's start with the number 0837: -8730-0378=8352 -8532-2358=6174 - - -6174 is called the Kaprekar constant. The process of sorting and subtracting and repeating this until either 0 or the Kaprekar constant is reached is called the Kaprekar routine. - - -We can consider the Kaprekar routine for other bases and number of digits. -Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers. -However, it can be shown that for 5 digits and a base b = 6t+39, a Kaprekar constant exists. -E.g. base 15: (10,4,14,9,5)15 + +6174 is a remarkable number; if we sort its digits in increasing order and subtract that number from the number you get when you sort the digits in decreasing order, we get 7641-1467=6174. +Even more remarkable is that if we start from any 4 digit number and repeat this process of sorting and subtracting, we'll eventually end up with 6174 or immediately with 0 if all digits are equal. +This also works with numbers that have less than 4 digits if we pad the number with leading zeroes until we have 4 digits. +E.g. let's start with the number 0837: +8730-0378=8352 +8532-2358=6174 + + +6174 is called the Kaprekar constant. The process of sorting and subtracting and repeating this until either 0 or the Kaprekar constant is reached is called the Kaprekar routine. + + +We can consider the Kaprekar routine for other bases and number of digits. +Unfortunately, it is not guaranteed a Kaprekar constant exists in all cases; either the routine can end up in a cycle for some input numbers or the constant the routine arrives at can be different for different input numbers. +However, it can be shown that for 5 digits and a base b = 6t+39, a Kaprekar constant exists. +E.g. base 15: (10,4,14,9,5)15 base 21: (14,6,20,13,7)21 - -Define Cb to be the Kaprekar constant in base b for 5 digits. -Define the function sb(i) to be - - 0 if i = Cb or if i written in base b consists of 5 identical digits - the number of iterations it takes the Kaprekar routine in base b to arrive at Cb, otherwise - -Note that we can define sb(i) for all integers i b5. If i written in base b takes less than 5 digits, the number is padded with leading zero digits until we have 5 digits before applying the Kaprekar routine. - - -Define S(b) as the sum of sb(i) for 0 i b5. -E.g. S(15) = 5274369 -S(111) = 400668930299 - - -Find the sum of S(6k+3) for 2 k 300. -Give the last 18 digits as your answer. + +Define Cb to be the Kaprekar constant in base b for 5 digits. +Define the function sb(i) to be + + 0 if i = Cb or if i written in base b consists of 5 identical digits + the number of iterations it takes the Kaprekar routine in base b to arrive at Cb, otherwise + +Note that we can define sb(i) for all integers i b5. If i written in base b takes less than 5 digits, the number is padded with leading zero digits until we have 5 digits before applying the Kaprekar routine. + + +Define S(b) as the sum of sb(i) for 0 i b5. +E.g. S(15) = 5274369 +S(111) = 400668930299 + + +Find the sum of S(6k+3) for 2 k 300. +Give the last 18 digits as your answer. diff --git a/project_euler/problems/415_titanic_sets.txt b/project_euler/problems/415_titanic_sets.txt index 212b312..1d8c088 100644 --- a/project_euler/problems/415_titanic_sets.txt +++ b/project_euler/problems/415_titanic_sets.txt @@ -5,7 +5,7 @@ Titanic sets A set of lattice points S is called a titanic set if there exists a line passing through exactly two points in S. An example of a titanic set is S = {(0, 0), (0, 1), (0, 2), (1, 1), (2, 0), (1, 0)}, where the line passing through (0, 1) and (2, 0) does not pass through any other point in S. On the other hand, the set {(0, 0), (1, 1), (2, 2), (4, 4)} is not a titanic set since the line passing through any two points in the set also passes through the other two. -For any positive integer N, let T(N) be the number of titanic sets S whose every point (x, y) satisfies 0 x, y N. +For any positive integer N, let T(N) be the number of titanic sets S whose every point (x, y) satisfies 0 x, y N. It can be verified that T(1) = 11, T(2) = 494, T(4) = 33554178, T(111) mod 108 = 13500401 and T(105) mod 108 = 63259062. Find T(1011) mod 108. diff --git a/project_euler/problems/416_a_frogs_trip.txt b/project_euler/problems/416_a_frogs_trip.txt index 31ac177..7d9cb35 100644 --- a/project_euler/problems/416_a_frogs_trip.txt +++ b/project_euler/problems/416_a_frogs_trip.txt @@ -3,7 +3,7 @@ http://projecteuler.net/problem=416 A frog's trip A row of n squares contains a frog in the leftmost square. By successive jumps the frog goes to the rightmost square and then back to the leftmost square. On the outward trip he jumps one, two or three squares to the right, and on the homeward trip he jumps to the left in a similar manner. He cannot jump outside the squares. He repeats the round-trip travel m times. -Let F(m, n) be the number of the ways the frog can travel so that at most one square remains unvisited. +Let F(m, n) be the number of the ways the frog can travel so that at most one square remains unvisited. For example, F(1, 3) = 4, F(1, 4) = 15, F(1, 5) = 46, F(2, 3) = 16 and F(2, 100) mod 109 = 429619151. Find the last 9 digits of F(10, 1012). diff --git a/project_euler/problems/417_reciprocal_cycles_ii.txt b/project_euler/problems/417_reciprocal_cycles_ii.txt index 1993658..e281f65 100644 --- a/project_euler/problems/417_reciprocal_cycles_ii.txt +++ b/project_euler/problems/417_reciprocal_cycles_ii.txt @@ -35,14 +35,14 @@ A unit fraction contains 1 in the numerator. The decimal representation of the u Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle. - -Unit fractions whose denominator has no other prime factors than 2 and/or 5 are not considered to have a recurring cycle. -We define the length of the recurring cycle of those unit fractions as 0. - -Let L(n) denote the length of the recurring cycle of 1/n. -You are given that L(n) for 3 n 1 000 000 equals 55535191115. +Unit fractions whose denominator has no other prime factors than 2 and/or 5 are not considered to have a recurring cycle. +We define the length of the recurring cycle of those unit fractions as 0. - -Find L(n) for 3 n 100 000 000 + +Let L(n) denote the length of the recurring cycle of 1/n. +You are given that L(n) for 3 n 1 000 000 equals 55535191115. + + +Find L(n) for 3 n 100 000 000 diff --git a/project_euler/problems/418_factorisation_triples.txt b/project_euler/problems/418_factorisation_triples.txt index 7ceb4f9..8df2eb1 100644 --- a/project_euler/problems/418_factorisation_triples.txt +++ b/project_euler/problems/418_factorisation_triples.txt @@ -2,18 +2,18 @@ http://projecteuler.net/problem=418 Factorisation triples - + Let n be a positive integer. An integer triple (a, b, c) is called a factorisation triple of n if: 1 a b c - a·b·c = n. + a·b·c = n. + + +Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n which minimises c / a. One can show that this triple is unique. + - -Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n which minimises c / a. One can show that this triple is unique. +For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872. - -For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872. - -Find f(43!). +Find f(43!). diff --git a/project_euler/problems/419_look_and_say_sequence.txt b/project_euler/problems/419_look_and_say_sequence.txt index 8ee53dd..cfa3e04 100644 --- a/project_euler/problems/419_look_and_say_sequence.txt +++ b/project_euler/problems/419_look_and_say_sequence.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=419 Look and say sequence - -The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... -The sequence starts with 1 and all other members are obtained by describing the previous member in terms of consecutive digits. -It helps to do this out loud: -1 is 'one one' 11 -11 is 'two ones' 21 -21 is 'one two and one one' 1211 -1211 is 'one one, one two and two ones' 111221 -111221 is 'three ones, two twos and one one' 312211 -... - - -Define A(n), B(n) and C(n) as the number of ones, twos and threes in the n'th element of the sequence respectively. -One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625. - - -Find A(n), B(n) and C(n) for n = 1012. -Give your answer modulo 230 and separate your values for A, B and C by a comma. -E.g. for n = 40 the answer would be 31254,20259,11625 + +The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... +The sequence starts with 1 and all other members are obtained by describing the previous member in terms of consecutive digits. +It helps to do this out loud: +1 is 'one one' 11 +11 is 'two ones' 21 +21 is 'one two and one one' 1211 +1211 is 'one one, one two and two ones' 111221 +111221 is 'three ones, two twos and one one' 312211 +... + + +Define A(n), B(n) and C(n) as the number of ones, twos and threes in the n'th element of the sequence respectively. +One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625. + + +Find A(n), B(n) and C(n) for n = 1012. +Give your answer modulo 230 and separate your values for A, B and C by a comma. +E.g. for n = 40 the answer would be 31254,20259,11625 diff --git a/project_euler/problems/420_2x2_positive_integer_matrix.txt b/project_euler/problems/420_2x2_positive_integer_matrix.txt index cb94194..93ea707 100644 --- a/project_euler/problems/420_2x2_positive_integer_matrix.txt +++ b/project_euler/problems/420_2x2_positive_integer_matrix.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=420 2x2 positive integer matrix -A positive integer matrix is a matrix whose elements are all positive integers. +A positive integer matrix is a matrix whose elements are all positive integers. Some positive integer matrices can be expressed as a square of a positive integer matrix in two different ways. Here is an example: - -We define F(N) as the number of the 2x2 positive integer matrices which have a trace less than N and which can be expressed as a square of a positive integer matrix in two different ways. -We can verify that F(50) = 7 and F(1000) = 1019. - -Find F(107). +We define F(N) as the number of the 2x2 positive integer matrices which have a trace less than N and which can be expressed as a square of a positive integer matrix in two different ways. +We can verify that F(50) = 7 and F(1000) = 1019. + + +Find F(107). diff --git a/project_euler/problems/421_prime_factors_of_n151.txt b/project_euler/problems/421_prime_factors_of_n151.txt index aad2e6e..aacae5a 100644 --- a/project_euler/problems/421_prime_factors_of_n151.txt +++ b/project_euler/problems/421_prime_factors_of_n151.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=421 Prime factors of n15+1 - -Numbers of the form n15+1 are composite for every integer n 1. -For positive integers n and m let s(n,m) be defined as the sum of the distinct prime factors of n15+1 not exceeding m. - -E.g. 215+1 = 3311331. -So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345. - -Also 1015+1 = 7111321124121619091. + +Numbers of the form n15+1 are composite for every integer n 1. +For positive integers n and m let s(n,m) be defined as the sum of the distinct prime factors of n15+1 not exceeding m. + +E.g. 215+1 = 3311331. +So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345. + +Also 1015+1 = 7111321124121619091. So s(10,100) = 31 and s(10,1000) = 483. - -Find ∑ s(n,108) for 1 n 1011. + +Find ∑ s(n,108) for 1 n 1011. diff --git a/project_euler/problems/422_sequence_of_points_on_a_hyperbola.txt b/project_euler/problems/422_sequence_of_points_on_a_hyperbola.txt index 075b375..35fb02f 100644 --- a/project_euler/problems/422_sequence_of_points_on_a_hyperbola.txt +++ b/project_euler/problems/422_sequence_of_points_on_a_hyperbola.txt @@ -4,11 +4,11 @@ Sequence of points on a hyperbola Let H be the hyperbola defined by the equation 12x2 + 7xy - 12y2 = 625. Next, define X as the point (7, 1). It can be seen that X is in H. -Now we define a sequence of points in H, {Pi : i 1}, as: +Now we define a sequence of points in H, {Pi : i 1}, as: - P1 = (13, 61/4). - P2 = (-43/6, -4). - For i > 2, Pi is the unique point in H that is different from Pi-1 and such that line PiPi-1 is parallel to line Pi-2X. It can be shown that Pi is well-defined, and that its coordinates are always rational. + P1 = (13, 61/4). + P2 = (-43/6, -4). + For i > 2, Pi is the unique point in H that is different from Pi-1 and such that line PiPi-1 is parallel to line Pi-2X. It can be shown that Pi is well-defined, and that its coordinates are always rational. You are given that P3 = (-19/2, -229/24), P4 = (1267/144, -37/12) and P7 = (17194218091/143327232, 274748766781/1719926784). diff --git a/project_euler/problems/423_consecutive_die_throws.txt b/project_euler/problems/423_consecutive_die_throws.txt index 3ad437f..a2004f1 100644 --- a/project_euler/problems/423_consecutive_die_throws.txt +++ b/project_euler/problems/423_consecutive_die_throws.txt @@ -2,16 +2,16 @@ http://projecteuler.net/problem=423 Consecutive die throws -Let n be a positive integer. +Let n be a positive integer. A 6-sided die is thrown n times. Let c be the number of pairs of consecutive throws that give the same value. -For example, if n = 7 and the values of the die throws are (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the same value: -(1,1,5,6,6,6,3) -(1,1,5,6,6,6,3) -(1,1,5,6,6,6,3) +For example, if n = 7 and the values of the die throws are (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the same value: +(1,1,5,6,6,6,3) +(1,1,5,6,6,6,3) +(1,1,5,6,6,6,3) Therefore, c = 3 for (1,1,5,6,6,6,3). -Define C(n) as the number of outcomes of throwing a 6-sided die n times such that c does not exceed π(n).1 +Define C(n) as the number of outcomes of throwing a 6-sided die n times such that c does not exceed π(n).1 For example, C(3) = 216, C(4) = 1290, C(11) = 361912500 and C(24) = 4727547363281250000. -Define S(L) as C(n) for 1 n L. +Define S(L) as C(n) for 1 n L. For example, S(50) mod 1 000 000 007 = 832833871. Find S(50 000 000) mod 1 000 000 007. 1 π denotes the prime-counting function, i.e. π(n) is the number of primes n. diff --git a/project_euler/problems/424_kakuro.txt b/project_euler/problems/424_kakuro.txt index e6e0ce1..d67abb3 100644 --- a/project_euler/problems/424_kakuro.txt +++ b/project_euler/problems/424_kakuro.txt @@ -5,35 +5,35 @@ Kakuro - -The above is an example of a cryptic kakuro (also known as cross sums, or even sums cross) puzzle, with its final solution on the right. (The common rules of kakuro puzzles can be found easily on numerous internet sites. Other related information can also be currently found at krazydad.com whose author has provided the puzzle data for this challenge.) - -The downloadable text file (kakuro200.txt) contains the description of 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file is the above example which is coded as follows: +The above is an example of a cryptic kakuro (also known as cross sums, or even sums cross) puzzle, with its final solution on the right. (The common rules of kakuro puzzles can be found easily on numerous internet sites. Other related information can also be currently found at krazydad.com whose author has provided the puzzle data for this challenge.) - -6,X,X,(vCC),(vI),X,X,X,(hH),B,O,(vCA),(vJE),X,(hFE,vD),O,O,O,O,(hA),O,I,(hJC,vB),O,O,(hJC),H,O,O,O,X,X,X,(hJE),O,O,X - -The first character is a numerical digit indicating the size of the information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7 (for a 6x6 puzzle) followed by a comma (,). The extra top line and left column are needed to insert information. +The downloadable text file (kakuro200.txt) contains the description of 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file is the above example which is coded as follows: - -The content of each cell is then described and followed by a comma, going left to right and starting with the top line. -X = Gray cell, not required to be filled by a digit. -O (upper case letter)= White empty cell to be filled by a digit. -A = Or any one of the upper case letters from A to J to be replaced by its equivalent digit in the solved puzzle. -( ) = Location of the encrypted sums. Horizontal sums are preceded by a lower case "h" and vertical sums are preceded by a lower case "v". Those are followed by one or two upper case letters depending if the sum is a single digit or double digit one. For double digit sums, the first letter would be for the "tens" and the second one for the "units". When the cell must contain information for both a horizontal and a vertical sum, the first one is always for the horizontal sum and the two are separated by a comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets is also immediately followed by a comma. - -The description of the last cell is followed by a Carriage Return/Line Feed (CRLF) instead of a comma. +6,X,X,(vCC),(vI),X,X,X,(hH),B,O,(vCA),(vJE),X,(hFE,vD),O,O,O,O,(hA),O,I,(hJC,vB),O,O,(hJC),H,O,O,O,X,X,X,(hJE),O,O,X - -The required answer to each puzzle is based on the value of each letter necessary to arrive at the solution and according to the alphabetical order. As indicated under the example puzzle, its answer would be 8426039571. At least 9 out of the 10 encrypting letters are always part of the problem description. When only 9 are given, the missing one must be assigned the remaining digit. - -You are given that the sum of the answers for the first 10 puzzles in the file is 64414157580. +The first character is a numerical digit indicating the size of the information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7 (for a 6x6 puzzle) followed by a comma (,). The extra top line and left column are needed to insert information. - -Find the sum of the answers for the 200 puzzles. + +The content of each cell is then described and followed by a comma, going left to right and starting with the top line. +X = Gray cell, not required to be filled by a digit. +O (upper case letter)= White empty cell to be filled by a digit. +A = Or any one of the upper case letters from A to J to be replaced by its equivalent digit in the solved puzzle. +( ) = Location of the encrypted sums. Horizontal sums are preceded by a lower case "h" and vertical sums are preceded by a lower case "v". Those are followed by one or two upper case letters depending if the sum is a single digit or double digit one. For double digit sums, the first letter would be for the "tens" and the second one for the "units". When the cell must contain information for both a horizontal and a vertical sum, the first one is always for the horizontal sum and the two are separated by a comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets is also immediately followed by a comma. + + +The description of the last cell is followed by a Carriage Return/Line Feed (CRLF) instead of a comma. + + +The required answer to each puzzle is based on the value of each letter necessary to arrive at the solution and according to the alphabetical order. As indicated under the example puzzle, its answer would be 8426039571. At least 9 out of the 10 encrypting letters are always part of the problem description. When only 9 are given, the missing one must be assigned the remaining digit. + + +You are given that the sum of the answers for the first 10 puzzles in the file is 64414157580. + + +Find the sum of the answers for the 200 puzzles. diff --git a/project_euler/problems/425_prime_connection.txt b/project_euler/problems/425_prime_connection.txt index eed51d9..262e9fa 100644 --- a/project_euler/problems/425_prime_connection.txt +++ b/project_euler/problems/425_prime_connection.txt @@ -2,24 +2,24 @@ http://projecteuler.net/problem=425 Prime connection - -Two positive numbers A and B are said to be connected (denoted by "A ↔ B") if one of these conditions holds: -(1) A and B have the same length and differ in exactly one digit; for example, 123 ↔ 173. -(2) Adding one digit to the left of A (or B) makes B (or A); for example, 23 ↔ 223 and 123 ↔ 23. - -We call a prime P a 2's relative if there exists a chain of connected primes between 2 and P and no prime in the chain exceeds P. +Two positive numbers A and B are said to be connected (denoted by "A ↔ B") if one of these conditions holds: +(1) A and B have the same length and differ in exactly one digit; for example, 123 ↔ 173. +(2) Adding one digit to the left of A (or B) makes B (or A); for example, 23 ↔ 223 and 123 ↔ 23. - -For example, 127 is a 2's relative. One of the possible chains is shown below: -2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127 -However, 11 and 103 are not 2's relatives. - -Let F(N) be the sum of the primes N which are not 2's relatives. -We can verify that F(103) = 431 and F(104) = 78728. +We call a prime P a 2's relative if there exists a chain of connected primes between 2 and P and no prime in the chain exceeds P. - -Find F(107). + +For example, 127 is a 2's relative. One of the possible chains is shown below: +2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127 +However, 11 and 103 are not 2's relatives. + + +Let F(N) be the sum of the primes N which are not 2's relatives. +We can verify that F(103) = 431 and F(104) = 78728. + + +Find F(107). diff --git a/project_euler/problems/426_boxball_system.txt b/project_euler/problems/426_boxball_system.txt index 58711de..7059478 100644 --- a/project_euler/problems/426_boxball_system.txt +++ b/project_euler/problems/426_boxball_system.txt @@ -2,37 +2,37 @@ http://projecteuler.net/problem=426 Box-ball system - -Consider an infinite row of boxes. Some of the boxes contain a ball. For example, an initial configuration of 2 consecutive occupied boxes followed by 2 empty boxes, 2 occupied boxes, 1 empty box, and 2 occupied boxes can be denoted by the sequence (2, 2, 2, 1, 2), in which the number of consecutive occupied and empty boxes appear alternately. - -A turn consists of moving each ball exactly once according to the following rule: Transfer the leftmost ball which has not been moved to the nearest empty box to its right. +Consider an infinite row of boxes. Some of the boxes contain a ball. For example, an initial configuration of 2 consecutive occupied boxes followed by 2 empty boxes, 2 occupied boxes, 1 empty box, and 2 occupied boxes can be denoted by the sequence (2, 2, 2, 1, 2), in which the number of consecutive occupied and empty boxes appear alternately. - -After one turn the sequence (2, 2, 2, 1, 2) becomes (2, 2, 1, 2, 3) as can be seen below; note that we begin the new sequence starting at the first occupied box. +A turn consists of moving each ball exactly once according to the following rule: Transfer the leftmost ball which has not been moved to the nearest empty box to its right. +After one turn the sequence (2, 2, 2, 1, 2) becomes (2, 2, 1, 2, 3) as can be seen below; note that we begin the new sequence starting at the first occupied box. - -A system like this is called a Box-Ball System or BBS for short. - -It can be shown that after a sufficient number of turns, the system evolves to a state where the consecutive numbers of occupied boxes is invariant. In the example below, the consecutive numbers of occupied boxes evolves to [1, 2, 3]; we shall call this the final state. + + + +A system like this is called a Box-Ball System or BBS for short. + + +It can be shown that after a sufficient number of turns, the system evolves to a state where the consecutive numbers of occupied boxes is invariant. In the example below, the consecutive numbers of occupied boxes evolves to [1, 2, 3]; we shall call this the final state. + - We define the sequence {ti}: -s0 = 290797 -sk+1 = sk2 mod 50515093 -tk = (sk mod 64) + 1 +s0 = 290797 +sk+1 = sk2 mod 50515093 +tk = (sk mod 64) + 1 + - -Starting from the initial configuration (t0, t1, …, t10), the final state becomes [1, 3, 10, 24, 51, 75]. -Starting from the initial configuration (t0, t1, …, t10 000 000), find the final state. -Give as your answer the sum of the squares of the elements of the final state. For example, if the final state is [1, 2, 3] then 14 ( = 12 + 22 + 32) is your answer. +Starting from the initial configuration (t0, t1, …, t10), the final state becomes [1, 3, 10, 24, 51, 75]. +Starting from the initial configuration (t0, t1, …, t10 000 000), find the final state. +Give as your answer the sum of the squares of the elements of the final state. For example, if the final state is [1, 2, 3] then 14 ( = 12 + 22 + 32) is your answer. diff --git a/project_euler/problems/427_nsequences.txt b/project_euler/problems/427_nsequences.txt index c0de70e..a0acfb8 100644 --- a/project_euler/problems/427_nsequences.txt +++ b/project_euler/problems/427_nsequences.txt @@ -2,9 +2,9 @@ http://projecteuler.net/problem=427 n-sequences -A sequence of integers S = {si} is called an n-sequence if it has n elements and each element si satisfies 1 si n. Thus there are nn distinct n-sequences in total. +A sequence of integers S = {si} is called an n-sequence if it has n elements and each element si satisfies 1 si n. Thus there are nn distinct n-sequences in total. For example, the sequence S = {1, 5, 5, 10, 7, 7, 7, 2, 3, 7} is a 10-sequence. -For any sequence S, let L(S) be the length of the longest contiguous subsequence of S with the same value. +For any sequence S, let L(S) be the length of the longest contiguous subsequence of S with the same value. For example, for the given sequence S above, L(S) = 3, because of the three consecutive 7's. Let f(n) = L(S) for all n-sequences S. For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121. diff --git a/project_euler/problems/428_necklace_of_circles.txt b/project_euler/problems/428_necklace_of_circles.txt index 0715b63..4fdc5a4 100644 --- a/project_euler/problems/428_necklace_of_circles.txt +++ b/project_euler/problems/428_necklace_of_circles.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=428 Necklace of circles -Let a, b and c be positive numbers. -Let W, X, Y, Z be four collinear points where |WX| = a, |XY| = b, |YZ| = c and |WZ| = a + b + c. -Let Cin be the circle having the diameter XY. +Let a, b and c be positive numbers. +Let W, X, Y, Z be four collinear points where |WX| = a, |XY| = b, |YZ| = c and |WZ| = a + b + c. +Let Cin be the circle having the diameter XY. Let Cout be the circle having the diameter WZ. - -The triplet (a, b, c) is called a necklace triplet if you can place k 3 distinct circles C1, C2, ..., Ck such that: -Ci has no common interior points with any Cj for 1 i, j k and i j, -Ci is tangent to both Cin and Cout for 1 i k, -Ci is tangent to Ci+1 for 1 i k, and -Ck is tangent to C1. +The triplet (a, b, c) is called a necklace triplet if you can place k 3 distinct circles C1, C2, ..., Ck such that: - -For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can be shown that (2, 2, 5) is not. +Ci has no common interior points with any Cj for 1 i, j k and i j, +Ci is tangent to both Cin and Cout for 1 i k, +Ci is tangent to Ci+1 for 1 i k, and +Ck is tangent to C1. - -Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c are positive integers, and b n. -For example, T(1) = 9, T(20) = 732 and T(3000) = 438106. +For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can be shown that (2, 2, 5) is not. - -Find T(1 000 000 000). + + +Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c are positive integers, and b n. +For example, T(1) = 9, T(20) = 732 and T(3000) = 438106. + + +Find T(1 000 000 000). diff --git a/project_euler/problems/429_sum_of_squares_of_unitary_divisors.txt b/project_euler/problems/429_sum_of_squares_of_unitary_divisors.txt index c3e84ca..1f5baef 100644 --- a/project_euler/problems/429_sum_of_squares_of_unitary_divisors.txt +++ b/project_euler/problems/429_sum_of_squares_of_unitary_divisors.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=429 Sum of squares of unitary divisors - -A unitary divisor d of a number n is a divisor of n that has the property gcd(d, n/d) = 1. -The unitary divisors of 4! = 24 are 1, 3, 8 and 24. -The sum of their squares is 12 + 32 + 82 + 242 = 650. - -Let S(n) represent the sum of the squares of the unitary divisors of n. Thus S(4!)=650. +A unitary divisor d of a number n is a divisor of n that has the property gcd(d, n/d) = 1. +The unitary divisors of 4! = 24 are 1, 3, 8 and 24. +The sum of their squares is 12 + 32 + 82 + 242 = 650. - -Find S(100 000 000!) modulo 1 000 000 009. + +Let S(n) represent the sum of the squares of the unitary divisors of n. Thus S(4!)=650. + + +Find S(100 000 000!) modulo 1 000 000 009. diff --git a/project_euler/problems/430_range_flips.txt b/project_euler/problems/430_range_flips.txt index fd4b9fb..1bd8067 100644 --- a/project_euler/problems/430_range_flips.txt +++ b/project_euler/problems/430_range_flips.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=430 Range flips -N disks are placed in a row, indexed 1 to N from left to right. +N disks are placed in a row, indexed 1 to N from left to right. Each disk has a black side and white side. Initially all disks show their white side. -At each turn, two, not necessarily distinct, integers A and B between 1 and N (inclusive) are chosen uniformly at random. +At each turn, two, not necessarily distinct, integers A and B between 1 and N (inclusive) are chosen uniformly at random. All disks with an index from A to B (inclusive) are flipped. The following example shows the case N = 8. At the first turn A = 5 and B = 2, and at the second turn A = 4 and B = 6. -Let E(N, M) be the expected number of disks that show their white side after M turns. +Let E(N, M) be the expected number of disks that show their white side after M turns. We can verify that E(3, 1) = 10/9, E(3, 2) = 5/3, E(10, 4) 5.157 and E(100, 10) 51.893. -Find E(1010, 4000). +Find E(1010, 4000). Give your answer rounded to 2 decimal places behind the decimal point. diff --git a/project_euler/problems/432_totient_sum.txt b/project_euler/problems/432_totient_sum.txt index 6a09f44..fe7e13a 100644 --- a/project_euler/problems/432_totient_sum.txt +++ b/project_euler/problems/432_totient_sum.txt @@ -2,12 +2,12 @@ http://projecteuler.net/problem=432 Totient sum - -Let S(n,m) = φ(n i) for 1 i m. (φ is Euler's totient function) -You are given that S(510510,106 )= 45480596821125120. - -Find S(510510,1011). -Give the last 9 digits of your answer. +Let S(n,m) = φ(n i) for 1 i m. (φ is Euler's totient function) +You are given that S(510510,106 )= 45480596821125120. + + +Find S(510510,1011). +Give the last 9 digits of your answer. diff --git a/project_euler/problems/433_steps_in_euclids_algorithm.txt b/project_euler/problems/433_steps_in_euclids_algorithm.txt index ba31def..a0f63de 100644 --- a/project_euler/problems/433_steps_in_euclids_algorithm.txt +++ b/project_euler/problems/433_steps_in_euclids_algorithm.txt @@ -2,20 +2,20 @@ http://projecteuler.net/problem=433 Steps in Euclid's algorithm - + Let E(x0, y0) be the number of steps it takes to determine the greatest common divisor of x0 and y0 with Euclid's algorithm. More formally: x1 = y0, y1 = x0 mod y0 -xn = yn-1, yn = xn-1 mod yn-1 -E(x0, y0) is the smallest n such that yn = 0. +xn = yn-1, yn = xn-1 mod yn-1 +E(x0, y0) is the smallest n such that yn = 0. + + +We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4. + - -We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4. +Define S(N) as the sum of E(x,y) for 1 x,y N. +We have S(1) = 1, S(10) = 221 and S(100) = 39826. - -Define S(N) as the sum of E(x,y) for 1 x,y N. -We have S(1) = 1, S(10) = 221 and S(100) = 39826. - -Find S(5·106). +Find S(5·106). diff --git a/project_euler/problems/434_rigid_graphs.txt b/project_euler/problems/434_rigid_graphs.txt index 6b824bd..cf1fa55 100644 --- a/project_euler/problems/434_rigid_graphs.txt +++ b/project_euler/problems/434_rigid_graphs.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=434 Rigid graphs -Recall that a graph is a collection of vertices and edges connecting the vertices, and that two vertices connected by an edge are called adjacent. -Graphs can be embedded in Euclidean space by associating each vertex with a point in the Euclidean space. -A flexible graph is an embedding of a graph where it is possible to move one or more vertices continuously so that the distance between at least two nonadjacent vertices is altered while the distances between each pair of adjacent vertices is kept constant. -A rigid graph is an embedding of a graph which is not flexible. -Informally, a graph is rigid if by replacing the vertices with fully rotating hinges and the edges with rods that are unbending and inelastic, no parts of the graph can be moved independently from the rest of the graph. +Recall that a graph is a collection of vertices and edges connecting the vertices, and that two vertices connected by an edge are called adjacent. +Graphs can be embedded in Euclidean space by associating each vertex with a point in the Euclidean space. +A flexible graph is an embedding of a graph where it is possible to move one or more vertices continuously so that the distance between at least two nonadjacent vertices is altered while the distances between each pair of adjacent vertices is kept constant. +A rigid graph is an embedding of a graph which is not flexible. +Informally, a graph is rigid if by replacing the vertices with fully rotating hinges and the edges with rods that are unbending and inelastic, no parts of the graph can be moved independently from the rest of the graph. The grid graphs embedded in the Euclidean plane are not rigid, as the following animation demonstrates: However, one can make them rigid by adding diagonal edges to the cells. For example, for the 2x3 grid graph, there are 19 ways to make the graph rigid: -Note that for the purposes of this problem, we do not consider changing the orientation of a diagonal edge or adding both diagonal edges to a cell as a different way of making a grid graph rigid. +Note that for the purposes of this problem, we do not consider changing the orientation of a diagonal edge or adding both diagonal edges to a cell as a different way of making a grid graph rigid. -Let R(m,n) be the number of ways to make the m n grid graph rigid. -E.g. R(2,3) = 19 and R(5,5) = 23679901 +Let R(m,n) be the number of ways to make the m n grid graph rigid. +E.g. R(2,3) = 19 and R(5,5) = 23679901 -Define S(N) as R(i,j) for 1 i, j N. -E.g. S(5) = 25021721. -Find S(100), give your answer modulo 1000000033 +Define S(N) as R(i,j) for 1 i, j N. +E.g. S(5) = 25021721. +Find S(100), give your answer modulo 1000000033 diff --git a/project_euler/problems/436_unfair_wager.txt b/project_euler/problems/436_unfair_wager.txt index bf7c403..683fe9f 100644 --- a/project_euler/problems/436_unfair_wager.txt +++ b/project_euler/problems/436_unfair_wager.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=436 Unfair wager -Julie proposes the following wager to her sister Louise. -She suggests they play a game of chance to determine who will wash the dishes. -For this game, they shall use a generator of independent random numbers uniformly distributed between 0 and 1. -The game starts with S = 0. -The first player, Louise, adds to S different random numbers from the generator until S > 1 and records her last random number 'x'. -The second player, Julie, continues adding to S different random numbers from the generator until S > 2 and records her last random number 'y'. +Julie proposes the following wager to her sister Louise. +She suggests they play a game of chance to determine who will wash the dishes. +For this game, they shall use a generator of independent random numbers uniformly distributed between 0 and 1. +The game starts with S = 0. +The first player, Louise, adds to S different random numbers from the generator until S > 1 and records her last random number 'x'. +The second player, Julie, continues adding to S different random numbers from the generator until S > 2 and records her last random number 'y'. The player with the highest number wins and the loser washes the dishes, i.e. if y > x the second player wins. -For example, if the first player draws 0.62 and 0.44, the first player turn ends since 0.62+0.44 > 1 and x = 0.44. -If the second players draws 0.1, 0.27 and 0.91, the second player turn ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91. +For example, if the first player draws 0.62 and 0.44, the first player turn ends since 0.62+0.44 > 1 and x = 0.44. +If the second players draws 0.1, 0.27 and 0.91, the second player turn ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91. Since y > x, the second player wins. -Louise thinks about it for a second, and objects: "That's not fair". -What is the probability that the second player wins? +Louise thinks about it for a second, and objects: "That's not fair". +What is the probability that the second player wins? Give your answer rounded to 10 places behind the decimal point in the form 0.abcdefghij diff --git a/project_euler/problems/437_fibonacci_primitive_roots.txt b/project_euler/problems/437_fibonacci_primitive_roots.txt index 10f01b8..a434ed9 100644 --- a/project_euler/problems/437_fibonacci_primitive_roots.txt +++ b/project_euler/problems/437_fibonacci_primitive_roots.txt @@ -2,26 +2,26 @@ http://projecteuler.net/problem=437 Fibonacci primitive roots - -When we calculate 8n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3, 2, 5, 7. -As we see all possible values from 1 to 10 occur. So 8 is a primitive root of 11. -But there is more: -If we take a closer look we see: -1+8=9 -8+9=176 mod 11 -9+6=154 mod 11 -6+4=10 -4+10=143 mod 11 -10+3=132 mod 11 -3+2=5 -2+5=7 -5+7=121 mod 11. - -So the powers of 8 mod 11 are cyclic with period 10, and 8n + 8n+1 ≡ 8n+2 (mod 11). -8 is called a Fibonacci primitive root of 11. -Not every prime has a Fibonacci primitive root. -There are 323 primes less than 10000 with one or more Fibonacci primitive roots and the sum of these primes is 1480491. -Find the sum of the primes less than 100,000,000 with at least one Fibonacci primitive root. - - + +When we calculate 8n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3, 2, 5, 7. +As we see all possible values from 1 to 10 occur. So 8 is a primitive root of 11. +But there is more: +If we take a closer look we see: +1+8=9 +8+9=176 mod 11 +9+6=154 mod 11 +6+4=10 +4+10=143 mod 11 +10+3=132 mod 11 +3+2=5 +2+5=7 +5+7=121 mod 11. + +So the powers of 8 mod 11 are cyclic with period 10, and 8n + 8n+1 ≡ 8n+2 (mod 11). +8 is called a Fibonacci primitive root of 11. +Not every prime has a Fibonacci primitive root. +There are 323 primes less than 10000 with one or more Fibonacci primitive roots and the sum of these primes is 1480491. +Find the sum of the primes less than 100,000,000 with at least one Fibonacci primitive root. + + diff --git a/project_euler/problems/438_integer_part_of_polynomial_equations_solutions.txt b/project_euler/problems/438_integer_part_of_polynomial_equations_solutions.txt index d601cec..8bb9d5c 100644 --- a/project_euler/problems/438_integer_part_of_polynomial_equations_solutions.txt +++ b/project_euler/problems/438_integer_part_of_polynomial_equations_solutions.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=438 Integer part of polynomial equation's solutions - -For an n-tuple of integers t = (a1, ..., an), let (x1, ..., xn) be the solutions of the polynomial equation xn + a1xn-1 + a2xn-2 + ... + an-1x + an = 0. - -Consider the following two conditions: +For an n-tuple of integers t = (a1, ..., an), let (x1, ..., xn) be the solutions of the polynomial equation xn + a1xn-1 + a2xn-2 + ... + an-1x + an = 0. -x1, ..., xn are all real. -If x1, ..., xn are sorted, xi = i for 1 i n. (·: floor function.) - -In the case of n = 4, there are 12 n-tuples of integers which satisfy both conditions. -We define S(t) as the sum of the absolute values of the integers in t. -For n = 4 we can verify that S(t) = 2087 for all n-tuples t which satisfy both conditions. +Consider the following two conditions: - -Find S(t) for n = 7. +x1, ..., xn are all real. +If x1, ..., xn are sorted, xi = i for 1 i n. (·: floor function.) + + +In the case of n = 4, there are 12 n-tuples of integers which satisfy both conditions. +We define S(t) as the sum of the absolute values of the integers in t. +For n = 4 we can verify that S(t) = 2087 for all n-tuples t which satisfy both conditions. + + +Find S(t) for n = 7. diff --git a/project_euler/problems/439_sum_of_sum_of_divisors.txt b/project_euler/problems/439_sum_of_sum_of_divisors.txt index a6fb087..882e8ba 100644 --- a/project_euler/problems/439_sum_of_sum_of_divisors.txt +++ b/project_euler/problems/439_sum_of_sum_of_divisors.txt @@ -2,9 +2,9 @@ http://projecteuler.net/problem=439 Sum of sum of divisors -Let d(k) be the sum of all divisors of k. -We define the function S(N) = 1iN 1jN d(i·j). +Let d(k) be the sum of all divisors of k. +We define the function S(N) = 1iN 1jN d(i·j). For example, S(3) = d(1) + d(2) + d(3) + d(2) + d(4) + d(6) + d(3) + d(6) + d(9) = 59. -You are given that S(103) = 563576517282 and S(105) mod 109 = 215766508. +You are given that S(103) = 563576517282 and S(105) mod 109 = 215766508. Find S(1011) mod 109. diff --git a/project_euler/problems/440_gcd_and_tiling.txt b/project_euler/problems/440_gcd_and_tiling.txt index d4b58ca..0316d8a 100644 --- a/project_euler/problems/440_gcd_and_tiling.txt +++ b/project_euler/problems/440_gcd_and_tiling.txt @@ -8,10 +8,10 @@ For example, here are some of the ways to tile a board of length n = 8: Let T(n) be the number of ways to tile a board of length n as described above. For example, T(1) = 10 and T(2) = 101. -Let S(L) be the triple sum a,b,c gcd(T(ca), T(cb)) for 1 a, b, c L. -For example: -S(2) = 10444 -S(3) = 1292115238446807016106539989 +Let S(L) be the triple sum a,b,c gcd(T(ca), T(cb)) for 1 a, b, c L. +For example: +S(2) = 10444 +S(3) = 1292115238446807016106539989 S(4) mod 987 898 789 = 670616280. Find S(2000) mod 987 898 789. diff --git a/project_euler/problems/441_the_inverse_summation_of_coprime_couples.txt b/project_euler/problems/441_the_inverse_summation_of_coprime_couples.txt index 7259f10..776bb48 100644 --- a/project_euler/problems/441_the_inverse_summation_of_coprime_couples.txt +++ b/project_euler/problems/441_the_inverse_summation_of_coprime_couples.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=441 The inverse summation of coprime couples - -For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer pairs p and q which satisfy all of these conditions: + +For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer pairs p and q which satisfy all of these conditions: 1 p q M p + q M p and q are coprime. - -We also define S(N) as the sum of R(i) for 2 i N. -We can verify that S(2) = R(2) = 1/2, S(10) 6.9147 and S(100) 58.2962. - -Find S(107). Give your answer rounded to four decimal places. +We also define S(N) as the sum of R(i) for 2 i N. +We can verify that S(2) = R(2) = 1/2, S(10) 6.9147 and S(100) 58.2962. + + +Find S(107). Give your answer rounded to four decimal places. diff --git a/project_euler/problems/443_gcd_sequence.txt b/project_euler/problems/443_gcd_sequence.txt index 8afa6bf..68f2778 100644 --- a/project_euler/problems/443_gcd_sequence.txt +++ b/project_euler/problems/443_gcd_sequence.txt @@ -2,8 +2,8 @@ http://projecteuler.net/problem=443 GCD sequence -Let g(n) be a sequence defined as follows: -g(4) = 13, +Let g(n) be a sequence defined as follows: +g(4) = 13, g(n) = g(n-1) + gcd(n, g(n-1)) for n 4. The first few values are: diff --git a/project_euler/problems/444_the_roundtable_lottery.txt b/project_euler/problems/444_the_roundtable_lottery.txt index 2ff0fba..f4eb249 100644 --- a/project_euler/problems/444_the_roundtable_lottery.txt +++ b/project_euler/problems/444_the_roundtable_lottery.txt @@ -4,12 +4,12 @@ The Roundtable Lottery A group of p people decide to sit down at a round table and play a lottery-ticket trading game. Each person starts off with a randomly-assigned, unscratched lottery ticket. Each ticket, when scratched, reveals a whole-pound prize ranging anywhere from £1 to £p, with no two tickets alike. The goal of the game is for each person to maximize his ticket winnings upon leaving the game. An arbitrary person is chosen to be the first player. Going around the table, each player has only one of two options: -1. The player can scratch his ticket and reveal its worth to everyone at the table. +1. The player can scratch his ticket and reveal its worth to everyone at the table. 2. The player can trade his unscratched ticket for a previous player's scratched ticket, and then leave the game with that ticket. The previous player then scratches his newly-acquired ticket and reveals its worth to everyone at the table. The game ends once all tickets have been scratched. All players still remaining at the table must leave with their currently-held tickets. Assume that each player uses the optimal strategy for maximizing the expected value of his ticket winnings. Let E(p) represent the expected number of players left at the table when the game ends in a game consisting of p players (e.g. E(111) = 5.2912 when rounded to 5 significant digits). -Let S1(N) = E(p) +Let S1(N) = E(p) Let Sk(N) = Sk-1(p) for k 1 Find S20(1014) and write the answer in scientific notation rounded to 10 significant digits. Use a lowercase e to separate mantissa and exponent (e.g. S3(100) = 5.983679014e5). diff --git a/project_euler/problems/445_retractions_a.txt b/project_euler/problems/445_retractions_a.txt index 7812075..eb28a3b 100644 --- a/project_euler/problems/445_retractions_a.txt +++ b/project_euler/problems/445_retractions_a.txt @@ -2,19 +2,19 @@ http://projecteuler.net/problem=445 Retractions A - -For every integer n1, the family of functions fn,a,b is defined -by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. -We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. -Let R(n) be the number of retractions for n. - +For every integer n1, the family of functions fn,a,b is defined +by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. +We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. +Let R(n) be the number of retractions for n. + + You are given that - R(c) for c=C(100 000,k), and 1 k 99 999 628701600 (mod 1 000 000 007). + R(c) for c=C(100 000,k), and 1 k 99 999 628701600 (mod 1 000 000 007). (C(n,k) is the binomial coefficient). - -Find R(c) for c=C(10 000 000,k), and 1 k 9 999 999. -Give your answer modulo 1 000 000 007. + +Find R(c) for c=C(10 000 000,k), and 1 k 9 999 999. +Give your answer modulo 1 000 000 007. diff --git a/project_euler/problems/446_retractions_b.txt b/project_euler/problems/446_retractions_b.txt index 312938b..ba1c540 100644 --- a/project_euler/problems/446_retractions_b.txt +++ b/project_euler/problems/446_retractions_b.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=446 Retractions B - -For every integer n1, the family of functions fn,a,b is defined -by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. -We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. -Let R(n) be the number of retractions for n. - - -F(N)=R(n4+4) for 1nN. + +For every integer n1, the family of functions fn,a,b is defined +by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. +We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. +Let R(n) be the number of retractions for n. + + +F(N)=R(n4+4) for 1nN. F(1024)=77532377300600. - -Find F(107) (mod 1 000 000 007) - + +Find F(107) (mod 1 000 000 007) + diff --git a/project_euler/problems/447_retractions_c.txt b/project_euler/problems/447_retractions_c.txt index eb85d1e..065fd05 100644 --- a/project_euler/problems/447_retractions_c.txt +++ b/project_euler/problems/447_retractions_c.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=447 Retractions C - -For every integer n1, the family of functions fn,a,b is defined -by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. -We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. -Let R(n) be the number of retractions for n. - - -F(N)=R(n) for 2nN. -F(107)638042271 (mod 1 000 000 007). - - -Find F(1014) (mod 1 000 000 007). + +For every integer n1, the family of functions fn,a,b is defined +by fn,a,b(x)ax+b mod n for a,b,x integer and 0an, 0bn, 0xn. +We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))fn,a,b(x) mod n for every 0xn. +Let R(n) be the number of retractions for n. + + +F(N)=R(n) for 2nN. +F(107)638042271 (mod 1 000 000 007). + + +Find F(1014) (mod 1 000 000 007). diff --git a/project_euler/problems/448_average_least_common_multiple.txt b/project_euler/problems/448_average_least_common_multiple.txt index 642a1e8..d91a9e6 100644 --- a/project_euler/problems/448_average_least_common_multiple.txt +++ b/project_euler/problems/448_average_least_common_multiple.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=448 Average least common multiple - -The function lcm(a,b) denotes the least common multiple of a and b. -Let A(n) be the average of the values of lcm(n,i) for 1in. -E.g: A(2)=(2+2)/2=2 and A(10)=(10+10+30+20+10+30+70+40+90+10)/10=32. - -Let S(n)=A(k) for 1kn. -S(100)=122726. - - -Find S(99999999019) mod 999999017. + +The function lcm(a,b) denotes the least common multiple of a and b. +Let A(n) be the average of the values of lcm(n,i) for 1in. +E.g: A(2)=(2+2)/2=2 and A(10)=(10+10+30+20+10+30+70+40+90+10)/10=32. + +Let S(n)=A(k) for 1kn. +S(100)=122726. + + +Find S(99999999019) mod 999999017. diff --git a/project_euler/problems/449_chocolate_covered_candy.txt b/project_euler/problems/449_chocolate_covered_candy.txt index 231101d..e7f03dc 100644 --- a/project_euler/problems/449_chocolate_covered_candy.txt +++ b/project_euler/problems/449_chocolate_covered_candy.txt @@ -2,22 +2,22 @@ http://projecteuler.net/problem=449 Chocolate covered candy -Phil the confectioner is making a new batch of chocolate covered candy. Each candy centre is shaped like an ellipsoid of revolution defined by the equation: b2x2 + b2y2 + a2z2 = a2b2. +Phil the confectioner is making a new batch of chocolate covered candy. Each candy centre is shaped like an ellipsoid of revolution defined by the equation: b2x2 + b2y2 + a2z2 = a2b2. - -Phil wants to know how much chocolate is needed to cover one candy centre with a uniform coat of chocolate one millimeter thick. -If a=1 mm and b=1 mm, the amount of chocolate required is - +Phil wants to know how much chocolate is needed to cover one candy centre with a uniform coat of chocolate one millimeter thick. + +If a=1 mm and b=1 mm, the amount of chocolate required is + 28 3 π mm3 - -If a=2 mm and b=1 mm, the amount of chocolate required is approximately 60.35475635 mm3. - - + +If a=2 mm and b=1 mm, the amount of chocolate required is approximately 60.35475635 mm3. + + Find the amount of chocolate in mm3 required if a=3 mm and b=1 mm. Give your answer as the number rounded to 8 decimal places behind the decimal point. diff --git a/project_euler/problems/450_hypocycloid_and_lattice_points.txt b/project_euler/problems/450_hypocycloid_and_lattice_points.txt index 5536d5e..c23623e 100644 --- a/project_euler/problems/450_hypocycloid_and_lattice_points.txt +++ b/project_euler/problems/450_hypocycloid_and_lattice_points.txt @@ -2,34 +2,34 @@ http://projecteuler.net/problem=450 Hypocycloid and Lattice points - -A hypocycloid is the curve drawn by a point on a small circle rolling inside a larger circle. The parametric equations of a hypocycloid centered at the origin, and starting at the right most point is given by: -$x(t) = (R - r) \cos(t) + r \cos(\frac {R - r} r t)$ -$y(t) = (R - r) \sin(t) - r \sin(\frac {R - r} r t)$ -Where R is the radius of the large circle and r the radius of the small circle. - +A hypocycloid is the curve drawn by a point on a small circle rolling inside a larger circle. The parametric equations of a hypocycloid centered at the origin, and starting at the right most point is given by: +$x(t) = (R - r) \cos(t) + r \cos(\frac {R - r} r t)$ +$y(t) = (R - r) \sin(t) - r \sin(\frac {R - r} r t)$ +Where R is the radius of the large circle and r the radius of the small circle. + + Let $C(R, r)$ be the set of distinct points with integer coordinates on the hypocycloid with radius R and r and for which there is a corresponding value of t such that $\sin(t)$ and $\cos(t)$ are rational numbers. - + Let $S(R, r) = \sum_{(x,y) \in C(R, r)} |x| + |y|$ be the sum of the absolute values of the x and y coordinates of the points in $C(R, r)$. - - -Let $T(N) = \sum_{R = 3}^N \sum_{r=1}^{\lfloor \frac {R - 1} 2 \rfloor} S(R, r)$ be the sum of $S(R, r)$ for R and r positive integers, $R\leq N$ and $2r - - + + +Let $T(N) = \sum_{R = 3}^N \sum_{r=1}^{\lfloor \frac {R - 1} 2 \rfloor} S(R, r)$ be the sum of $S(R, r)$ for R and r positive integers, $R\leq N$ and $2r + + You are given: C(3, 1) = {(3, 0), (-1, 2), (-1,0), (-1,-2)} C(2500, 1000) = - - {(2500, 0), (772, 2376), (772, -2376), (516, 1792), + + {(2500, 0), (772, 2376), (772, -2376), (516, 1792), (516, -1792), (500, 0), (68, 504), (68, -504),(-1356, 1088), (-1356, -1088), (-1500, 1000), (-1500, -1000)} Note: (-625, 0) is not an element of C(2500, 1000) because $\sin(t)$ is not a rational number for the corresponding values of t. S(3, 1) = (|3| + |0|) + (|-1| + |2|) + (|-1| + |0|) + (|-1| + |-2|) = 10 -T(3) = 10; T(10) = 524 ;T(100) = 580442; T(103) = 583108600. +T(3) = 10; T(10) = 524 ;T(100) = 580442; T(103) = 583108600. + - -Find T(106). +Find T(106). diff --git a/project_euler/problems/451_modular_inverses.txt b/project_euler/problems/451_modular_inverses.txt index 9740053..a1a87f9 100644 --- a/project_euler/problems/451_modular_inverses.txt +++ b/project_euler/problems/451_modular_inverses.txt @@ -2,23 +2,23 @@ http://projecteuler.net/problem=451 Modular inverses - -Consider the number 15. -There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14. -The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 -because -1*1 mod 15=1 -2*8=16 mod 15=1 -4*4=16 mod 15=1 -7*13=91 mod 15=1 -11*11=121 mod 15=1 + +Consider the number 15. +There are eight positive numbers less than 15 which are coprime to 15: 1, 2, 4, 7, 8, 11, 13, 14. +The modular inverses of these numbers modulo 15 are: 1, 8, 4, 13, 2, 11, 7, 14 +because +1*1 mod 15=1 +2*8=16 mod 15=1 +4*4=16 mod 15=1 +7*13=91 mod 15=1 +11*11=121 mod 15=1 14*14=196 mod 15=1 - -Let I(n) be the largest positive number m smaller than n-1 such that the modular inverse of m modulo n equals m itself. -So I(15)=11. + +Let I(n) be the largest positive number m smaller than n-1 such that the modular inverse of m modulo n equals m itself. +So I(15)=11. Also I(100)=51 and I(7)=1. - + Find I(n) for 3n2·107 diff --git a/project_euler/problems/454_diophantine_reciprocals_iii.txt b/project_euler/problems/454_diophantine_reciprocals_iii.txt index ecdf14f..4287a3a 100644 --- a/project_euler/problems/454_diophantine_reciprocals_iii.txt +++ b/project_euler/problems/454_diophantine_reciprocals_iii.txt @@ -15,6 +15,6 @@ In the following equation x, y, and n are positive integers. For a limit L we define F(L) as the number of solutions which satisfy x y L. -We can verify that F(15) = 4 and F(1000) = 1069. +We can verify that F(15) = 4 and F(1000) = 1069. Find F(1012). diff --git a/project_euler/problems/456_triangles_containing_the_origin_ii.txt b/project_euler/problems/456_triangles_containing_the_origin_ii.txt index d412807..d42f059 100644 --- a/project_euler/problems/456_triangles_containing_the_origin_ii.txt +++ b/project_euler/problems/456_triangles_containing_the_origin_ii.txt @@ -4,17 +4,17 @@ Triangles containing the origin II Define: xn = (1248n mod 32323) - 16161 -yn = (8421n mod 30103) - 15051 -Pn = {(x1, y1), (x2, y2), ..., (xn, yn)} +yn = (8421n mod 30103) - 15051 +Pn = {(x1, y1), (x2, y2), ..., (xn, yn)} For example, P8 = {(-14913, -6630), (-10161, 5625), (5226, 11896), (8340, -10778), (15852, -5203), (-15165, 11295), (-1427, -14495), (12407, 1060)}. Let C(n) be the number of triangles whose vertices are in Pn which contain the origin in the interior. - -Examples: -C(8) = 20 -C(600) = 8950634 -C(40 000) = 2666610948988 -Find C(2 000 000). - +Examples: +C(8) = 20 +C(600) = 8950634 +C(40 000) = 2666610948988 + +Find C(2 000 000). + diff --git a/project_euler/problems/457_a_polynomial_modulo_the_square_of_a_prime.txt b/project_euler/problems/457_a_polynomial_modulo_the_square_of_a_prime.txt index dda717a..c769c3a 100644 --- a/project_euler/problems/457_a_polynomial_modulo_the_square_of_a_prime.txt +++ b/project_euler/problems/457_a_polynomial_modulo_the_square_of_a_prime.txt @@ -2,15 +2,15 @@ http://projecteuler.net/problem=457 A polynomial modulo the square of a prime - -Let f(n) = n2 - 3n - 1. -Let p be a prime. -Let R(p) be the smallest positive integer n such that f(n) mod p2 = 0 if such an integer n exists, otherwise R(p) = 0. - -Let SR(L) be ∑R(p) for all primes not exceeding L. +Let f(n) = n2 - 3n - 1. +Let p be a prime. +Let R(p) be the smallest positive integer n such that f(n) mod p2 = 0 if such an integer n exists, otherwise R(p) = 0. - -Find SR(107). + +Let SR(L) be ∑R(p) for all primes not exceeding L. + + +Find SR(107). diff --git a/project_euler/problems/458_permutations_of_project.txt b/project_euler/problems/458_permutations_of_project.txt index b90efe2..ad572a9 100644 --- a/project_euler/problems/458_permutations_of_project.txt +++ b/project_euler/problems/458_permutations_of_project.txt @@ -2,13 +2,13 @@ http://projecteuler.net/problem=458 Permutations of Project - -Consider the alphabet A made out of the letters of the word "project": A={c,e,j,o,p,r,t}. -Let T(n) be the number of strings of length n consisting of letters from A that do not have a substring that is one of the 5040 permutations of "project". - -T(7)=77-7!=818503. - - -Find T(1012). Give the last 9 digits of your answer. + +Consider the alphabet A made out of the letters of the word "project": A={c,e,j,o,p,r,t}. +Let T(n) be the number of strings of length n consisting of letters from A that do not have a substring that is one of the 5040 permutations of "project". + +T(7)=77-7!=818503. + + +Find T(1012). Give the last 9 digits of your answer. diff --git a/project_euler/problems/459_flipping_game.txt b/project_euler/problems/459_flipping_game.txt index c4e2607..8aff214 100644 --- a/project_euler/problems/459_flipping_game.txt +++ b/project_euler/problems/459_flipping_game.txt @@ -2,10 +2,10 @@ http://projecteuler.net/problem=459 Flipping game -The flipping game is a two player game played on a N by N square board. -Each square contains a disk with one side white and one side black. +The flipping game is a two player game played on a N by N square board. +Each square contains a disk with one side white and one side black. The game starts with all disks showing their white side. -A turn consists of flipping all disks in a rectangle with the following properties: +A turn consists of flipping all disks in a rectangle with the following properties: the upper right corner of the rectangle contains a white disk the rectangle width is a perfect square (1, 4, 9, 16, ...) @@ -13,7 +13,7 @@ the rectangle height is a triangular number (1, 3, 6, 10, ...) Players alternate turns. A player wins by turning the grid all black. -Let W(N) be the number of winning moves for the first player on a N by N board with all disks white, assuming perfect play. +Let W(N) be the number of winning moves for the first player on a N by N board with all disks white, assuming perfect play. W(1) = 1, W(2) = 0, W(5) = 8 and W(102) = 31395. For N=5, the first player's eight winning first moves are: diff --git a/project_euler/problems/460_an_ant_on_the_move.txt b/project_euler/problems/460_an_ant_on_the_move.txt index f82bcdc..1f604fc 100644 --- a/project_euler/problems/460_an_ant_on_the_move.txt +++ b/project_euler/problems/460_an_ant_on_the_move.txt @@ -2,29 +2,29 @@ http://projecteuler.net/problem=460 An ant on the move - -On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d. - -In each step, the ant at point (x0, y0) chooses one of the lattice points (x1, y1) which satisfy x1 0 and y1 1 and goes straight to (x1, y1) at a constant velocity v. The value of v depends on y0 and y1 as follows: +On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d. + + +In each step, the ant at point (x0, y0) chooses one of the lattice points (x1, y1) which satisfy x1 0 and y1 1 and goes straight to (x1, y1) at a constant velocity v. The value of v depends on y0 and y1 as follows: If y0 = y1, the value of v equals y0. If y0 y1, the value of v equals (y1 - y0) / (ln(y1) - ln(y0)). - -The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P1(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) 1.8205. Then the required time is sqrt(5) / 1.8205 1.2283. -From P1(1, 3) to P2(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 0.6667. From P2(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) 1.8205 so the required time is sqrt(5) / 1.8205 1.2283. -Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233. - -The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4. +The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P1(1, 3) at velocity (3 - 1) / (ln(3) - ln(1)) 1.8205. Then the required time is sqrt(5) / 1.8205 1.2283. +From P1(1, 3) to P2(3, 3) the ant travels at velocity 3 so the required time is 2 / 3 0.6667. From P2(3, 3) to B(4, 1) the ant travels at velocity (1 - 3) / (ln(1) - ln(3)) 1.8205 so the required time is sqrt(5) / 1.8205 1.2283. +Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233. + + +The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4. + + +Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) 2.960516287. +We can verify that F(10) 4.668187834 and F(100) 9.217221972. - -Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) 2.960516287. -We can verify that F(10) 4.668187834 and F(100) 9.217221972. - -Find F(10000). Give your answer rounded to nine decimal places. +Find F(10000). Give your answer rounded to nine decimal places. diff --git a/project_euler/problems/461_almost_pi.txt b/project_euler/problems/461_almost_pi.txt index 736545d..25804d8 100644 --- a/project_euler/problems/461_almost_pi.txt +++ b/project_euler/problems/461_almost_pi.txt @@ -5,7 +5,7 @@ Almost Pi Let fn(k) = ek/n - 1, for all non-negative integers k. Remarkably, f200(6) + f200(75) + f200(89) + f200(226) = 3.141592644529…  π. In fact, it is the best approximation of π of the form fn(a) + fn(b) + fn(c) + fn(d) for n = 200. -Let g(n) = a2 + b2 + c2 + d 2 for a, b, c, d that minimize the error: | fn(a) + fn(b) + fn(c) + fn(d) - π| +Let g(n) = a2 + b2 + c2 + d 2 for a, b, c, d that minimize the error: | fn(a) + fn(b) + fn(c) + fn(d) - π| (where |x| denotes the absolute value of x). You are given g(200) = 62 + 752 + 892 + 2262 = 64658. Find g(10000). diff --git a/project_euler/problems/462_permutation_of_3smooth_numbers.txt b/project_euler/problems/462_permutation_of_3smooth_numbers.txt index e385347..8ae9a31 100644 --- a/project_euler/problems/462_permutation_of_3smooth_numbers.txt +++ b/project_euler/problems/462_permutation_of_3smooth_numbers.txt @@ -2,21 +2,21 @@ http://projecteuler.net/problem=462 Permutation of 3-smooth numbers - -A 3-smooth number is an integer which has no prime factor larger than 3. For an integer N, we define S(N) as the set of 3-smooth numbers less than or equal to N . For example, S(20) = { 1, 2, 3, 4, 6, 8, 9, 12, 16, 18 }. - - -We define F(N) as the number of permutations of S(N) in which each element comes after all of its proper divisors. - - -This is one of the possible permutations for N = 20. -- 1, 2, 4, 3, 9, 8, 16, 6, 18, 12. -This is not a valid permutation because 12 comes before its divisor 6. -- 1, 2, 4, 3, 9, 8, 12, 16, 6, 18. - - -We can verify that F(6) = 5, F(8) = 9, F(20) = 450 and F(1000) 8.8521816557e21. -Find F(1018). Give as your answer its scientific notation rounded to ten digits after the decimal point. -When giving your answer, use a lowercase e to separate mantissa and exponent. E.g. if the answer is 112,233,445,566,778,899 then the answer format would be 1.1223344557e17. + +A 3-smooth number is an integer which has no prime factor larger than 3. For an integer N, we define S(N) as the set of 3-smooth numbers less than or equal to N . For example, S(20) = { 1, 2, 3, 4, 6, 8, 9, 12, 16, 18 }. + + +We define F(N) as the number of permutations of S(N) in which each element comes after all of its proper divisors. + + +This is one of the possible permutations for N = 20. +- 1, 2, 4, 3, 9, 8, 16, 6, 18, 12. +This is not a valid permutation because 12 comes before its divisor 6. +- 1, 2, 4, 3, 9, 8, 12, 16, 6, 18. + + +We can verify that F(6) = 5, F(8) = 9, F(20) = 450 and F(1000) 8.8521816557e21. +Find F(1018). Give as your answer its scientific notation rounded to ten digits after the decimal point. +When giving your answer, use a lowercase e to separate mantissa and exponent. E.g. if the answer is 112,233,445,566,778,899 then the answer format would be 1.1223344557e17. diff --git a/project_euler/problems/463_a_weird_recurrence_relation.txt b/project_euler/problems/463_a_weird_recurrence_relation.txt index be60a18..8b5a4b4 100644 --- a/project_euler/problems/463_a_weird_recurrence_relation.txt +++ b/project_euler/problems/463_a_weird_recurrence_relation.txt @@ -2,14 +2,14 @@ http://projecteuler.net/problem=463 A weird recurrence relation - -The function $f$ is defined for all positive integers as follows: -$f(1)=1$ -$f(3)=3$ -$f(2n)=f(n)$ -$f(4n + 1)=2f(2n + 1) - f(n)$ -$f(4n + 3)=3f(2n + 1) - 2f(n)$ +The function $f$ is defined for all positive integers as follows: + +$f(1)=1$ +$f(3)=3$ +$f(2n)=f(n)$ +$f(4n + 1)=2f(2n + 1) - f(n)$ +$f(4n + 3)=3f(2n + 1) - 2f(n)$ The function $S(n)$ is defined as $\sum_{i=1}^{n}f(i)$. diff --git a/project_euler/problems/464_mbius_function_and_intervals.txt b/project_euler/problems/464_mbius_function_and_intervals.txt index d19e2e5..e14b8ee 100644 --- a/project_euler/problems/464_mbius_function_and_intervals.txt +++ b/project_euler/problems/464_mbius_function_and_intervals.txt @@ -2,28 +2,28 @@ http://projecteuler.net/problem=464 Möbius function and intervals - -The Möbius function, denoted μ(n), is defined as: + +The Möbius function, denoted μ(n), is defined as: μ(n) = (-1)ω(n) if n is squarefree (where ω(n) is the number of distinct prime factors of n) μ(n) = 0 if n is not squarefree. - -Let P(a,b) be the number of integers n in the interval [a,b] such that μ(n) = 1. -Let N(a,b) be the number of integers n in the interval [a,b] such that μ(n) = -1. -For example, P(2,10) = 2 and N(2,10) = 4. - -Let C(n) be the number of integer pairs (a,b) such that: +Let P(a,b) be the number of integers n in the interval [a,b] such that μ(n) = 1. +Let N(a,b) be the number of integers n in the interval [a,b] such that μ(n) = -1. +For example, P(2,10) = 2 and N(2,10) = 4. + + +Let C(n) be the number of integer pairs (a,b) such that: 1  a  b  n, 99·N(a,b)  100·P(a,b), and 99·P(a,b)  100·N(a,b). - -For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319. - -Find C(20 000 000). +For example, C(10) = 13, C(500) = 16676 and C(10 000) = 20155319. + + +Find C(20 000 000). diff --git a/project_euler/problems/465_polar_polygons.txt b/project_euler/problems/465_polar_polygons.txt index 34cab5b..7823356 100644 --- a/project_euler/problems/465_polar_polygons.txt +++ b/project_euler/problems/465_polar_polygons.txt @@ -4,7 +4,7 @@ Polar polygons The kernel of a polygon is defined by the set of points from which the entire polygon's boundary is visible. We define a polar polygon as a polygon for which the origin is strictly contained inside its kernel. For this problem, a polygon can have collinear consecutive vertices. However, a polygon still cannot have self-intersection and cannot have zero area. -For example, only the first of the following is a polar polygon (the kernels of the second, third, and fourth do not strictly contain the origin, and the fifth does not have a kernel at all): +For example, only the first of the following is a polar polygon (the kernels of the second, third, and fourth do not strictly contain the origin, and the fifth does not have a kernel at all): Notice that the first polygon has three consecutive collinear vertices. Let P(n) be the number of polar polygons such that the vertices (x, y) have integer coordinates whose absolute values are not greater than n. diff --git a/project_euler/problems/466_distinct_terms_in_a_multiplication_table.txt b/project_euler/problems/466_distinct_terms_in_a_multiplication_table.txt index 49d1233..32821ac 100644 --- a/project_euler/problems/466_distinct_terms_in_a_multiplication_table.txt +++ b/project_euler/problems/466_distinct_terms_in_a_multiplication_table.txt @@ -2,26 +2,26 @@ http://projecteuler.net/problem=466 Distinct terms in a multiplication table - -table.p466, table.p466 th, table.p466 td { - border-width: 1px 1px 1px 1px; - border-style: solid solid solid solid; - border-color: black black black black; - text-align:right; - -moz-border-radius: 0px 0px 0px 0px; -} -table.p466 { - table-layout: fixed; - border-spacing: 1px; - border-collapse: separate; - background-color: rgb(224,237,252); -} -table.p466 th, table.p466 td { - padding: 1px 6px 1px 6px; - width: 20%; -} -table.p466 th { background-color: rgb(193,218,249); } -table.p466 td { background-color: rgb(255,255,255); } + +table.p466, table.p466 th, table.p466 td { + border-width: 1px 1px 1px 1px; + border-style: solid solid solid solid; + border-color: black black black black; + text-align:right; + -moz-border-radius: 0px 0px 0px 0px; +} +table.p466 { + table-layout: fixed; + border-spacing: 1px; + border-collapse: separate; + background-color: rgb(224,237,252); +} +table.p466 th, table.p466 td { + padding: 1px 6px 1px 6px; + width: 20%; +} +table.p466 th { background-color: rgb(193,218,249); } +table.p466 td { background-color: rgb(255,255,255); } Let P(m,n) be the number of distinct terms in an mn multiplication table. For example, a 34 multiplication table looks like this: @@ -32,9 +32,9 @@ For example, a 34 multiplication table looks like this: 3 36912 There are 8 distinct terms {1,2,3,4,6,8,9,12}, therefore P(3,4) = 8. -You are given that: -P(64,64) = 1263, -P(12,345) = 1998, and +You are given that: +P(64,64) = 1263, +P(12,345) = 1998, and P(32,1015) = 13826382602124302. Find P(64,1016). diff --git a/project_euler/problems/467_superinteger.txt b/project_euler/problems/467_superinteger.txt index 47b26bb..4666da1 100644 --- a/project_euler/problems/467_superinteger.txt +++ b/project_euler/problems/467_superinteger.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=467 Superinteger -An integer s is called a superinteger of another integer n if the digits of n form a subsequence of the digits of s. -For example, 2718281828 is a superinteger of 18828, while 314159 is not a superinteger of 151. +An integer s is called a superinteger of another integer n if the digits of n form a subsequence of the digits of s. +For example, 2718281828 is a superinteger of 18828, while 314159 is not a superinteger of 151. -Let p(n) be the nth prime number, and let c(n) be the nth composite number. For example, p(1) = 2, p(10) = 29, c(1) = 4 and c(10) = 18. -{p(i) : i 1} = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...} +Let p(n) be the nth prime number, and let c(n) be the nth composite number. For example, p(1) = 2, p(10) = 29, c(1) = 4 and c(10) = 18. +{p(i) : i 1} = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...} {c(i) : i 1} = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, ...} -Let PD the sequence of the digital roots of {p(i)} (CD is defined similarly for {c(i)}): -PD = {2, 3, 5, 7, 2, 4, 8, 1, 5, 2, ...} +Let PD the sequence of the digital roots of {p(i)} (CD is defined similarly for {c(i)}): +PD = {2, 3, 5, 7, 2, 4, 8, 1, 5, 2, ...} CD = {4, 6, 8, 9, 1, 3, 5, 6, 7, 9, ...} -Let Pn be the integer formed by concatenating the first n elements of PD (Cn is defined similarly for CD). -P10 = 2357248152 +Let Pn be the integer formed by concatenating the first n elements of PD (Cn is defined similarly for CD). +P10 = 2357248152 C10 = 4689135679 Let f(n) be the smallest positive integer that is a common superinteger of Pn and Cn. For example, f(10) = 2357246891352679, and f(100) mod 1 000 000 007 = 771661825. Find f(10 000) mod 1 000 000 007. diff --git a/project_euler/problems/468_smooth_divisors_of_binomial_coefficients.txt b/project_euler/problems/468_smooth_divisors_of_binomial_coefficients.txt index ba763b6..37f8acc 100644 --- a/project_euler/problems/468_smooth_divisors_of_binomial_coefficients.txt +++ b/project_euler/problems/468_smooth_divisors_of_binomial_coefficients.txt @@ -3,15 +3,15 @@ http://projecteuler.net/problem=468 Smooth divisors of binomial coefficients An integer is called B-smooth if none of its prime factors is greater than B. -Let SB(n) be the largest B-smooth divisor of n. -Examples: -S1(10) = 1 -S4(2100) = 12 +Let SB(n) be the largest B-smooth divisor of n. +Examples: +S1(10) = 1 +S4(2100) = 12 S17(2496144) = 5712 -Define F(n) = 1Bn 0rn SB(C(n,r)). Here, C(n,r) denotes the binomial coefficient. -Examples: -F(11) = 3132 -F(1 111) mod 1 000 000 993 = 706036312 +Define F(n) = 1Bn 0rn SB(C(n,r)). Here, C(n,r) denotes the binomial coefficient. +Examples: +F(11) = 3132 +F(1 111) mod 1 000 000 993 = 706036312 F(111 111) mod 1 000 000 993 = 22156169 Find F(11 111 111) mod 1 000 000 993. diff --git a/project_euler/problems/469_empty_chairs.txt b/project_euler/problems/469_empty_chairs.txt index 0e39f78..60b5567 100644 --- a/project_euler/problems/469_empty_chairs.txt +++ b/project_euler/problems/469_empty_chairs.txt @@ -2,17 +2,17 @@ http://projecteuler.net/problem=469 Empty chairs - -In a room N chairs are placed around a round table. -Knights enter the room one by one and choose at random an available empty chair. -To have enough elbow room the knights always leave at least one empty chair between each other. - - -When there aren't any suitable chairs left, the fraction C of empty chairs is determined. -We also define E(N) as the expected value of C. -We can verify that E(4) = 1/2 and E(6) = 5/9. - - -Find E(1018). Give your answer rounded to fourteen decimal places in the form 0.abcdefghijklmn. + +In a room N chairs are placed around a round table. +Knights enter the room one by one and choose at random an available empty chair. +To have enough elbow room the knights always leave at least one empty chair between each other. + + +When there aren't any suitable chairs left, the fraction C of empty chairs is determined. +We also define E(N) as the expected value of C. +We can verify that E(4) = 1/2 and E(6) = 5/9. + + +Find E(1018). Give your answer rounded to fourteen decimal places in the form 0.abcdefghijklmn.