fix(curriculum): format challenge markdown as per linting rules (#36326)

* fix: Format challenge markdown according to linting rules

* fix: Put spaces after section tags

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-221-alexandrian-integers.english.md

@@ -6,22 +6,11 @@ title: 'Problem 221: Alexandrian Integers'
 
 ## Description
 <section id='description'>
+
 We shall call a positive integer A an "Alexandrian integer", if there exist integers p, q, r such that:
+A = p · q · r and 1/A = 1/p + 1/q + 1/r
+<!-- TODO Use MathJax -->
 
-
-      A = p · q · r    and  
-
-
-      1A
-   =
-
-      1p
-   +
-
-      1q
-   +
-
-      1r
 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.
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-228-minkowski-sums.english.md

@@ -6,16 +6,17 @@ title: 'Problem 228: Minkowski Sums'
 
 ## Description
 <section id='description'>
+
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
 Let Sn be the regular n-sided polygon – or shape – whose vertices
 
 vk (k = 1,2,…,n) have coordinates:
 
-    xk   =  
-        cos( 2k-1/n ×180° )
+xk = cos( 2k-1/n ×180° )
 
-    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.
+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
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-301-nim.english.md

@@ -6,16 +6,17 @@ title: 'Problem 301: Nim'
 
 ## Description
 <section id='description'>
+
 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.
 - 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:
-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:
+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)

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-330-eulers-number.english.md

@@ -6,45 +6,24 @@ title: 'Problem 330: Euler''s Number'
 
 ## Description
 <section id='description'>
+
 An infinite sequence of real numbers a(n) is defined for all integers n as follows:
-
-
-For example,a(0) =
-    11!
-    +
-    12!
-    +
-    13!
-    + ... = e − 1
-a(1) =
-    e − 11!
-    +
-    12!
-    +
-    13!
-    + ... = 2e − 3
-a(2) =
-    2e − 31!
-    +
-    e − 12!
-    +
-    13!
-    + ... =
-    72
-    e − 6
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
+For example,a(0) = 11! + 12! + 13! + ... = e − 1
+a(1) = e − 11! + 12! + 13! + ... = 2e − 3
+a(2) = 2e − 31! + e − 12! + 13! + ... = 72 e − 6
 
 with e = 2.7182818... being Euler's constant.
 
 
 It can be shown that a(n) is of the form
 
-    A(n) e + B(n)n!
-    for integers A(n) and B(n).
+A(n) e + B(n)n!
+for integers A(n) and B(n).
 
 For example a(10) =
 
-    328161643 e − 65269448610!
-    .
+328161643 e − 65269448610!.
 
 Find A(109) + B(109) and give your answer mod 77 777 777.
 </section>

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-334-spilling-the-beans.english.md

@@ -6,6 +6,7 @@ title: 'Problem 334: Spilling the beans'
 
 ## Description
 <section id='description'>
+
 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.
@@ -13,49 +14,17 @@ A child plays a game, which allows only one kind of move: removing two beans fro
 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:
 
 
-
+<!-- TODO Use MathJax and re-write from projecteuler.net -->
 You are given the following sequences:
-      t0 = 123456.
-
-
-      ti =
-
-
-
-
-
-         ti-12
-
-         ,
-
-
-
-         if ti-1 is even
-
-
-
-         ti-12
-
-
-
-         926252,
-
-
-         if ti-1 is odd
-
-
-
-
-
-      where ⌊x⌋ is the floor function
-
-
-
-
-      and  is the bitwise XOR operator.
-
-
-      bi = ( ti mod 211) + 1.
+t0 = 123456.
+ti = ti-12,
+if ti-1 is even
+ti-12
+926252,
+if ti-1 is odd
+where ⌊x⌋ is the floor function
+and  is the bitwise XOR operator.
+bi = ( ti mod 211) + 1.
 
 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.

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-385-ellipses-inside-triangles.english.md

@@ -6,13 +6,16 @@ title: 'Problem 385: Ellipses inside triangles'
 
 ## Description
 <section id='description'>
+
 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.
 
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-443-gcd-sequence.english.md

@@ -6,15 +6,16 @@ title: 'Problem 443: GCD sequence'
 
 ## Description
 <section id='description'>
+
 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:
 
-    n4567891011121314151617181920...
-    g(n)1314161718272829303132333451545560...
-
+n 4567891011121314151617181920...
+g(n) 1314161718272829303132333451545560...
+<!-- TODO Use MathJax -->
 
 You are given that g(1 000) = 2524 and g(1 000 000) = 2624152.
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-449-chocolate-covered-candy.english.md

@@ -6,14 +6,16 @@ title: 'Problem 449: Chocolate covered candy'
 
 ## Description
 <section id='description'>
+
 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
 
-    283
-    π mm3
+283
+π mm3
+<!-- TODO Use MathJax -->
 
 If a=2 mm and b=1 mm, the amount of chocolate required is approximately 60.35475635 mm3.
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-454-diophantine-reciprocals-iii.english.md

@@ -6,13 +6,10 @@ title: 'Problem 454: Diophantine reciprocals III'
 
 ## Description
 <section id='description'>
-In the following equation x, y, and n are positive integers.
 
-1x
- +
-1y
- =
-1n
+In the following equation x, y, and n are positive integers.
+1/x + 1/y= 1/n
+<!-- TODO Use MathJax -->
 
 For a limit L we define F(L) as the number of solutions which satisfy x < y ≤ L.
 

curriculum/challenges/english/08-coding-interview-prep/project-euler/problem-462-permutation-of-3-smooth-numbers.english.md

@@ -6,6 +6,8 @@ title: 'Problem 462: Permutation of 3-smooth numbers'
 
 ## Description
 <section id='description'>
+
+<!-- TODO use MathJax -->
 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 }.
 
 
@@ -13,10 +15,12 @@ We define F(N) as the number of permutations of S(N) in which each element comes
 
 
 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.
 
+- 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.