remove $ from title field of module to allow CI passing

src/P108/problem.jl

@@ -14,18 +14,18 @@ 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:
 
-$$\begin{align}
+\begin{align}
 \dfrac{1}{5} + \dfrac{1}{20} &= \dfrac{1}{4}\\
 \dfrac{1}{6} + \dfrac{1}{12} &= \dfrac{1}{4}\\
 \dfrac{1}{8} + \dfrac{1}{8} &= \dfrac{1}{4}
 \end{align}
-$$
+
 
 
 What is the least value of 

src/P110/problem.jl

@@ -13,9 +13,9 @@ y
 , and 
 n
  are positive integers.
-$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$
-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?
+\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 
 Problem 108
  and as it is well beyond the limitations of a brute force approach it requires a clever implementation.

src/P137/problem.jl

@@ -6,31 +6,31 @@ 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 $k$th 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.
+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
-$\begin{align*} 
+\begin{align*} 
 A_F(\tfrac{1}{2})
  &= (\tfrac{1}{2})\times 1 + (\tfrac{1}{2})^2\times 1 + (\tfrac{1}{2})^3\times 2 + (\tfrac{1}{2})^4\times 3 + (\tfrac{1}{2})^5\times 5 + \cdots \\ 
  &= \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{2}{8} + \tfrac{3}{16} + \tfrac{5}{32} + \cdots \\
  &= 2
-\end{align*}$
+\end{align*}
 The corresponding values of 
 x
  for the first five natural numbers are shown below.
-$x$
-$A_F(x)$
-$\sqrt{2}-1$
+x
+A_F(x)
+\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.
+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

@@ -6,10 +6,10 @@ published_on = "Saturday, 20th January 2007, 11:00 am"
 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.
+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.
 """
 
 

src/P140/problem.jl

@@ -6,22 +6,22 @@ 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 $k$th 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.
-$x$
-$A_G(x)$
-$\frac{\sqrt{5}-1}{4}$
+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}
 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.
+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:
-$$\begin{align}\dfrac{1}{2} &= \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} +\\
+\begin{align}\dfrac{1}{2} &= \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + \dfrac{1}{5^2} +\\
 &\quad \dfrac{1}{7^2} + \dfrac{1}{12^2} + \dfrac{1}{15^2} + \dfrac{1}{20^2} +\\
-&\quad \dfrac{1}{28^2} + \dfrac{1}{35^2}\end{align}$$
+&\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

@@ -6,19 +6,19 @@ published_on = "Saturday, 3rd May 2008, 05:00 am"
 solved_by = 1673
 difficulty_rating = "75%"
 content = """
-Let $x$ be a real number.
+Let x be a real number.
 
 A 
 best approximation
- to $x$ for the 
+ 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 $6$.
-We shall call a real number $x$ 
+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 $10^8$?
+, 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?
 """
 
 

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

@@ -15,10 +15,10 @@ q
 , 
 r
  such that:
-$$A = p \cdot q \cdot r$$
+A = p \cdot q \cdot r
 and
-$$\dfrac{1}{A} = \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r}$$
-For example, 630 is an Alexandrian integer ($p = 5, q = -7, r = -18$).
+\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
  Alexandrian integer,  the first 6 Alexandrian integers being: 6, 42, 120, 156, 420, and 630.

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

@@ -24,28 +24,28 @@ 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 $x_{k+1} = x_k$.
+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.
+Since x_2 = x_1, 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…).
 
@@ -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