Merge pull request #5 from udohjeremiah/invalid-escape-sequences

edit files to solve invalid escape sequence errors

src/P108/problem.jl

@@ -14,17 +14,17 @@ y
 n
  are positive integers.
 
-\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}
+\\dfrac{1}{x} + \\dfrac{1}{y} = \\dfrac{1}{n}
 
 For 
 n
  = 4 there are exactly three distinct solutions:
 
-\begin{align}
-\dfrac{1}{5} + \dfrac{1}{20} &= \dfrac{1}{4}\\
-\dfrac{1}{6} + \dfrac{1}{12} &= \dfrac{1}{4}\\
-\dfrac{1}{8} + \dfrac{1}{8} &= \dfrac{1}{4}
-\end{align}
+\\begin{align}
+\\dfrac{1}{5} + \\dfrac{1}{20} &= \\dfrac{1}{4}\\
+\\dfrac{1}{6} + \\dfrac{1}{12} &= \\dfrac{1}{4}\\
+\\dfrac{1}{8} + \\dfrac{1}{8} &= \\dfrac{1}{4}
+\\end{align}
 
 
 

src/P110/problem.jl

@@ -13,7 +13,7 @@ y
 , and 
 n
  are positive integers.
-\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}
+\\dfrac{1}{x} + \\dfrac{1}{y} = \\dfrac{1}{n}
 It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for which the total number of distinct solutions exceeds one hundred.
 What is the least value of n for which the number of distinct solutions exceeds four million?
 NOTE: This problem is a much more difficult version of 

src/P137/problem.jl

@@ -6,29 +6,29 @@ published_on = "Friday, 12th January 2007, 06:00 pm"
 solved_by = 5717
 difficulty_rating = "50%"
 content = """
-Consider the infinite polynomial series A_F(x) = x F_1 + x^2 F_2 + x^3 F_3 + \dots, where F_k is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, \dots; that is, F_k = F_{k-1} + F_{k-2}, F_1 = 1 and F_2 = 1.
+Consider the infinite polynomial series A_F(x) = x F_1 + x^2 F_2 + x^3 F_3 + \\dots, where F_k is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, \\dots; that is, F_k = F_{k-1} + F_{k-2}, F_1 = 1 and F_2 = 1.
 For this problem we shall be interested in values of x for which A_F(x) is a positive integer.
 Surprisingly
-\begin{align*} 
-A_F(\tfrac{1}{2})
- &= (\tfrac{1}{2})\times 1 + (\tfrac{1}{2})^2\times 1 + (\tfrac{1}{2})^3\times 2 + (\tfrac{1}{2})^4\times 3 + (\tfrac{1}{2})^5\times 5 + \cdots \\ 
- &= \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{2}{8} + \tfrac{3}{16} + \tfrac{5}{32} + \cdots \\
+\\begin{align*} 
+A_F(\\tfrac{1}{2})
+ &= (\\tfrac{1}{2})\\times 1 + (\\tfrac{1}{2})^2\\times 1 + (\\tfrac{1}{2})^3\\times 2 + (\\tfrac{1}{2})^4\\times 3 + (\\tfrac{1}{2})^5\\times 5 + \\cdots \\ 
+ &= \\tfrac{1}{2} + \\tfrac{1}{4} + \\tfrac{2}{8} + \\tfrac{3}{16} + \\tfrac{5}{32} + \\cdots \\
  &= 2
-\end{align*}
+\\end{align*}
 The corresponding values of 
 x
  for the first five natural numbers are shown below.
 x
 A_F(x)
-\sqrt{2}-1
+\\sqrt{2}-1
 1
-\tfrac{1}{2}
+\\tfrac{1}{2}
 2
-\frac{\sqrt{13}-2}{3}
+\\frac{\\sqrt{13}-2}{3}
 3
-\frac{\sqrt{89}-5}{8}
+\\frac{\\sqrt{89}-5}{8}
 4
-\frac{\sqrt{34}-3}{5}
+\\frac{\\sqrt{34}-3}{5}
 5
 We shall call A_F(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.
 Find the 15th golden nugget.

src/P138/problem.jl

@@ -7,9 +7,9 @@ solved_by = 6029
 difficulty_rating = "45%"
 content = """
 Consider the isosceles triangle with base length, b = 16, and legs, L = 17.
-By using the Pythagorean theorem it can be seen that the height of the triangle, h = \sqrt{17^2 - 8^2} = 15, which is one less than the base length.
-With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b \pm 1.
-Find \sum L for the twelve smallest isosceles triangles for which h = b \pm 1 and b, L are positive integers.
+By using the Pythagorean theorem it can be seen that the height of the triangle, h = \\sqrt{17^2 - 8^2} = 15, which is one less than the base length.
+With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b \\pm 1.
+Find \\sum L for the twelve smallest isosceles triangles for which h = b \\pm 1 and b, L are positive integers.
 """
 
 

src/P140/problem.jl

@@ -6,20 +6,20 @@ published_on = "Saturday, 3rd February 2007, 07:00 am"
 solved_by = 4493
 difficulty_rating = "55%"
 content = """
-Consider the infinite polynomial series A_G(x) = x G_1 + x^2 G_2 + x^3 G_3 + \cdots, where G_k is the kth term of the second order recurrence relation G_k = G_{k-1} + G_{k-2}, G_1 = 1 and G_2 = 4; that is, 1, 4, 5, 9, 14, 23, \dots .
+Consider the infinite polynomial series A_G(x) = x G_1 + x^2 G_2 + x^3 G_3 + \\cdots, where G_k is the kth term of the second order recurrence relation G_k = G_{k-1} + G_{k-2}, G_1 = 1 and G_2 = 4; that is, 1, 4, 5, 9, 14, 23, \\dots .
 For this problem we shall be concerned with values of x for which A_G(x) is a positive integer.
 The corresponding values of x for the first five natural numbers are shown below.
 x
 A_G(x)
-\frac{\sqrt{5}-1}{4}
+\\frac{\\sqrt{5}-1}{4}
 1
-\tfrac{2}{5}
+\\tfrac{2}{5}
 2
-\frac{\sqrt{22}-2}{6}
+\\frac{\\sqrt{22}-2}{6}
 3
-\frac{\sqrt{137}-5}{14}
+\\frac{\\sqrt{137}-5}{14}
 4
-\tfrac{1}{2}
+\\tfrac{1}{2}
 5
 We shall call A_G(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365.
 Find the sum of the first thirty golden nuggets.

src/P152/problem.jl

@@ -10,9 +10,9 @@ There are several ways to write the number 1/2 as a sum of inverse squares using
 distinct
  integers.
 For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:
-\begin{align}\dfrac{1}{2} &= \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} +\\
-&\quad \dfrac{1}{7^2} + \dfrac{1}{12^2} + \dfrac{1}{15^2} + \dfrac{1}{20^2} +\\
-&\quad \dfrac{1}{28^2} + \dfrac{1}{35^2}\end{align}
+\\begin{align}\\dfrac{1}{2} &= \\dfrac{1}{2^2} + \\dfrac{1}{3^2} + \\dfrac{1}{4^2} + \\dfrac{1}{5^2} +\\
+&\\quad \\dfrac{1}{7^2} + \\dfrac{1}{12^2} + \\dfrac{1}{15^2} + \\dfrac{1}{20^2} +\\
+&\\quad \\dfrac{1}{28^2} + \\dfrac{1}{35^2}\\end{align}
 In fact, only using integers between 2 and 45 inclusive, there are exactly three ways to do it, the remaining two being: {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}.
 How many ways are there to write the number 1/2 as a sum of inverse squares using distinct integers between 2 and 80 inclusive?
 """

src/P153/problem.jl

@@ -91,7 +91,7 @@ n
 
 If for example we divide 5 by 1+2
 i
- we can simplify \dfrac{5}{1 + 2i} in the following manner:
+ we can simplify \\dfrac{5}{1 + 2i} in the following manner:
 
 
 Multiply numerator and denominator by the complex conjugate of 1+2
@@ -101,7 +101,7 @@ i
 .
 
 
-The result is \dfrac{5}{1 + 2i} = \dfrac{5}{1 + 2i}\dfrac{1 - 2i}{1 - 2i} = \dfrac{5(1 - 2i)}{1 - (2i)^2} = \dfrac{5(1 - 2i)}{1 - (-4)} = \dfrac{5(1 - 2i)}{5} = 1 - 2i.
+The result is \\dfrac{5}{1 + 2i} = \\dfrac{5}{1 + 2i}\\dfrac{1 - 2i}{1 - 2i} = \\dfrac{5(1 - 2i)}{1 - (2i)^2} = \\dfrac{5(1 - 2i)}{1 - (-4)} = \\dfrac{5(1 - 2i)}{5} = 1 - 2i.
 
 
 So 1+2
@@ -111,7 +111,7 @@ i
 
 Note that 1+
 i
- is not a divisor of 5 because \dfrac{5}{1 + i} = \dfrac{5}{2} - \dfrac{5}{2}i.
+ is not a divisor of 5 because \\dfrac{5}{1 + i} = \\dfrac{5}{2} - \\dfrac{5}{2}i.
 
 
 Note also that if the Gaussian Integer (
@@ -184,9 +184,9 @@ i
 i
 , 5
 12
-For divisors with positive real parts, then, we have: \sum \limits_{n = 1}^{5} {s(n)} = 35.
-For \sum \limits_{n = 1}^{10^5} {s(n)} = 17924657155.
-What is \sum \limits_{n = 1}^{10^8} {s(n)}?
+For divisors with positive real parts, then, we have: \\sum \\limits_{n = 1}^{5} {s(n)} = 35.
+For \\sum \\limits_{n = 1}^{10^5} {s(n)} = 17924657155.
+What is \\sum \\limits_{n = 1}^{10^8} {s(n)}?
 """
 
 

src/P155/problem.jl

@@ -22,7 +22,7 @@ Find D(18).
 
 Reminder : When connecting capacitors C₁, C₂ etc in parallel, the total capacitance is C_T= C₁ + C₂ +...,
 
-whereas when connecting them in series, the overall capacitance is given by: \dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + ...
+whereas when connecting them in series, the overall capacitance is given by: \\dfrac{1}{C_T} = \\dfrac{1}{C_1} + \\dfrac{1}{C_2} + ...
 """
 
 

src/P192/problem.jl

@@ -12,13 +12,13 @@ A
 best approximation
  to x for the 
 denominator bound
- d is a rational number \frac r s   in
+ d is a rational number \\frac r s   in
  reduced form
-, with s \le d, such that any rational number which is closer to x than \frac r s has a denominator larger than d:
- |\frac p q -x | < |\frac r s -x| \Rightarrow q > d
+, with s \\le d, such that any rational number which is closer to x than \\frac r s has a denominator larger than d:
+ |\\frac p q -x | < |\\frac r s -x| \\Rightarrow q > d
 
-For example, the best approximation to \sqrt {13} for the denominator bound 20 is \frac {18} 5 and the best approximation to \sqrt {13} for the denominator bound 30 is \frac {101}{28}.
-Find the sum of all denominators of the best approximations to \sqrt n for the denominator bound 10^{12}, where n is not a perfect square and  1 < n \le 100000. 
+For example, the best approximation to \\sqrt {13} for the denominator bound 20 is \\frac {18} 5 and the best approximation to \\sqrt {13} for the denominator bound 30 is \\frac {101}{28}.
+Find the sum of all denominators of the best approximations to \\sqrt n for the denominator bound 10^{12}, where n is not a perfect square and  1 < n \\le 100000. 
 """
 
 

src/P198/problem.jl

@@ -6,12 +6,12 @@ published_on = "Saturday, 14th June 2008, 02:00 am"
 solved_by = 1154
 difficulty_rating = "80%"
 content = """
-A best approximation to a real number x for the denominator bound d is a rational number \frac r s (in reduced form) with s \le d, so that any rational number \frac p q which is closer to x than \frac 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. \frac 9 {40} has the two best approximations \frac 1 4 and \frac 1 5 for the denominator bound 6.
+A best approximation to a real number x for the denominator bound d is a rational number \\frac r s (in reduced form) with s \\le d, so that any rational number \\frac p q which is closer to x than \\frac 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. \\frac 9 {40} has the two best approximations \\frac 1 4 and \\frac 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=\frac p q, 0 < x < \frac 1 {100}, are there whose denominator q does not exceed 10^8?
+How many ambiguous numbers x=\\frac p q, 0 < x < \\frac 1 {100}, are there whose denominator q does not exceed 10^8?
 """
 
 

src/P203/problem.jl

@@ -6,7 +6,7 @@ published_on = "Saturday, 6th September 2008, 02:00 pm"
 solved_by = 9350
 difficulty_rating = "25%"
 content = """
-The binomial coefficients \displaystyle \binom n k can be arranged in triangular form, Pascal's triangle, like this:
+The binomial coefficients \\displaystyle \\binom n k can be arranged in triangular form, Pascal's triangle, like this:
 1
 1
 1

src/P221/problem.jl

@@ -15,9 +15,9 @@ q
 , 
 r
  such that:
-A = p \cdot q \cdot r
+A = p \\cdot q \\cdot r
 and
-\dfrac{1}{A} = \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r}
+\\dfrac{1}{A} = \\dfrac{1}{p} + \\dfrac{1}{q} + \\dfrac{1}{r}
 For example, 630 is an Alexandrian integer (p = 5, q = -7, r = -18).
 In fact, 630 is the 6
 th

src/P226/problem.jl

@@ -8,11 +8,11 @@ difficulty_rating = "65%"
 content = """
 The 
 blancmange curve
- is the set of points (x, y) such that 0 \le x \le 1 and y = \sum \limits_{n = 0}^{\infty} {\dfrac{s(2^n x)}{2^n}}, where s(x) is the distance from x to the nearest integer.
+ is the set of points (x, y) such that 0 \\le x \\le 1 and y = \\sum \\limits_{n = 0}^{\\infty} {\\dfrac{s(2^n x)}{2^n}}, where s(x) is the distance from x to the nearest integer.
 The area under the blancmange curve is equal to ½, shown in pink in the diagram below.
 Let 
 C
- be the circle with centre \left ( \frac{1}{4}, \frac{1}{2} \right ) and radius \frac{1}{4}, shown in black in the diagram.
+ be the circle with centre \\left ( \\frac{1}{4}, \\frac{1}{2} \\right ) and radius \\frac{1}{4}, shown in black in the diagram.
 What area under the blancmange curve is enclosed by 
 C
 ?

src/P231/problem.jl

@@ -6,14 +6,14 @@ published_on = "Friday, 6th February 2009, 01:00 pm"
 solved_by = 5565
 difficulty_rating = "40%"
 content = """
-The binomial coefficient \displaystyle \binom {10} 3 = 120.
+The binomial coefficient \\displaystyle \\binom {10} 3 = 120.
 
-120 = 2^3 \times 3 \times 5 = 2 \times 2 \times 2 \times 3 \times 5, and 2 + 2 + 2 + 3 + 5 = 14.
+120 = 2^3 \\times 3 \\times 5 = 2 \\times 2 \\times 2 \\times 3 \\times 5, and 2 + 2 + 2 + 3 + 5 = 14.
 
-So the sum of the terms in the prime factorisation of \displaystyle \binom {10} 3 is 14.
+So the sum of the terms in the prime factorisation of \\displaystyle \\binom {10} 3 is 14.
 
 
-Find the sum of the terms in the prime factorisation of \displaystyle \binom {20\,000\,000} {15\,000\,000}.
+Find the sum of the terms in the prime factorisation of \\displaystyle \\binom {20\\,000\\,000} {15\\,000\\,000}.
 
 """
 

src/P241/problem.jl

@@ -6,12 +6,12 @@ published_on = "Saturday, 18th April 2009, 02:00 am"
 solved_by = 953
 difficulty_rating = "80%"
 content = """
-For a positive integer n, let \sigma(n) be the sum of all divisors of n. For example, \sigma(6) = 1 + 2 + 3 + 6 = 12.
-A perfect number, as you probably know, is a number with \sigma(n) = 2n.
+For a positive integer n, let \\sigma(n) be the sum of all divisors of n. For example, \\sigma(6) = 1 + 2 + 3 + 6 = 12.
+A perfect number, as you probably know, is a number with \\sigma(n) = 2n.
 Let us define the 
 perfection quotient
- of a positive integer as p(n) = \dfrac{\sigma(n)}{n}.
-Find the sum of all positive integers n \le 10^{18} for which p(n) has the form k + \dfrac{1}{2}, where k is an integer.
+ of a positive integer as p(n) = \\dfrac{\\sigma(n)}{n}.
+Find the sum of all positive integers n \\le 10^{18} for which p(n) has the form k + \\dfrac{1}{2}, where k is an integer.
 """
 
 

src/P245/problem.jl

@@ -7,15 +7,15 @@ solved_by = 829
 difficulty_rating = "80%"
 content = """
 We shall call a fraction that cannot be cancelled down a resilient fraction.
- Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient; for example, R(12) = \dfrac{4}{11}.
-The resilience of a number d \gt 1 is then \dfrac{\varphi(d)}{d - 1}, where \varphi is Euler's totient function.
+ Furthermore we shall define the resilience of a denominator, R(d), to be the ratio of its proper fractions that are resilient; for example, R(12) = \\dfrac{4}{11}.
+The resilience of a number d \\gt 1 is then \\dfrac{\\varphi(d)}{d - 1}, where \\varphi is Euler's totient function.
 We further define the 
 coresilience
- of a number n \gt 1 as C(n) = \dfrac{n - \varphi(n)}{n - 1}.
-The coresilience of a prime p is C(p) = \dfrac{1}{p - 1}.
+ of a number n \\gt 1 as C(n) = \\dfrac{n - \\varphi(n)}{n - 1}.
+The coresilience of a prime p is C(p) = \\dfrac{1}{p - 1}.
 Find the sum of all 
 composite
- integers 1 \lt n \le 2 \times 10^{11}, for which C(n) is a 
+ integers 1 \\lt n \\le 2 \\times 10^{11}, for which C(n) is a 
 unit fraction
 .
 """

src/P251/problem.jl

@@ -15,7 +15,7 @@ b
 c
 ) is called a Cardano Triplet if it satisfies the condition:
 
-\sqrt[3]{a + b \sqrt{c}} + \sqrt[3]{a - b \sqrt{c}} = 1
+\\sqrt[3]{a + b \\sqrt{c}} + \\sqrt[3]{a - b \\sqrt{c}} = 1
 
 
 

src/P255/problem.jl

@@ -24,26 +24,26 @@ n
 
 If 
 d
- is odd, set x_0 = 2 \times 10^{(d-1)/2}.
+ is odd, set x_0 = 2 \\times 10^{(d-1)/2}.
 
 If 
 d
- is even, set x_0 = 7 \times 10^{(d-2)/2}.
+ is even, set x_0 = 7 \\times 10^{(d-2)/2}.
 
 Repeat:
 
-x_{k+1} = \Biggl\lfloor{\dfrac{x_k + \lceil{n / x_k}\rceil}{2}}\Biggr\rfloor
+x_{k+1} = \\Biggl\\lfloor{\\dfrac{x_k + \\lceil{n / x_k}\\rceil}{2}}\\Biggr\\rfloor
 
 
 until x_{k+1} = x_k.
 As an example, let us find the rounded-square-root of 
 n
  = 4321.
 n
- has 4 digits, so x_0 = 7 \times 10^{(4-2)/2} = 70.
+ has 4 digits, so x_0 = 7 \\times 10^{(4-2)/2} = 70.
 
-x_1 = \Biggl\lfloor{\dfrac{70 + \lceil{4321 / 70}\rceil}{2}}\Biggr\rfloor = 66
-x_2 = \Biggl\lfloor{\dfrac{66 + \lceil{4321 / 66}\rceil}{2}}\Biggr\rfloor = 66
+x_1 = \\Biggl\\lfloor{\\dfrac{70 + \\lceil{4321 / 70}\\rceil}{2}}\\Biggr\\rfloor = 66
+x_2 = \\Biggl\\lfloor{\\dfrac{66 + \\lceil{4321 / 66}\\rceil}{2}}\\Biggr\\rfloor = 66
 
 Since x_2 = x_1, we stop here.
 
@@ -65,7 +65,7 @@ n
 
 Give your answer rounded to 10 decimal places.
 
-Note: The symbols \lfloor x \rfloor and \lceil x \rceil represent the 
+Note: The symbols \\lfloor x \\rfloor and \\lceil x \\rceil represent the 
 floor function
  and 
 ceiling function

src/P277/problem.jl

@@ -9,20 +9,20 @@ content = """
 
 A modified Collatz sequence of integers is obtained from a starting value a_1 in the following way:
 
-a_{n+1} = \, \,\, \frac {a_n} 3 \quad if a_n is divisible by 3. We shall denote this as a large downward step, "D".
+a_{n+1} = \\, \\,\\, \\frac {a_n} 3 \\quad if a_n is divisible by 3. We shall denote this as a large downward step, "D".
 
-a_{n+1} = \frac {4 a_n+2} 3 \, \, if a_n divided by 3 gives a remainder of 1. We shall denote this as an upward step, "U".
+a_{n+1} = \\frac {4 a_n+2} 3 \\, \\, if a_n divided by 3 gives a remainder of 1. We shall denote this as an upward step, "U".
 
 
-a_{n+1} = \frac {2 a_n -1} 3 \, \, if a_n divided by 3 gives a remainder of 2. We shall denote this as a small downward step, "d".
+a_{n+1} = \\frac {2 a_n -1} 3 \\, \\, if a_n divided by 3 gives a remainder of 2. We shall denote this as a small downward step, "d".
 
 
 The sequence terminates when some a_n = 1.
 
 
 Given any integer, we can list out the sequence of steps.
 
-For instance if a_1=231, then the sequence \{a_n\}=\{231,77,51,17,11,7,10,14,9,3,1\} corresponds to the steps "DdDddUUdDD".
+For instance if a_1=231, then the sequence \\{a_n\\}=\\{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....".