Feat: add new Markdown parser (#39800)

and change all the challenges to new `md` format.

curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-461-almost-pi.md

@@ -1,42 +1,37 @@
 ---
 id: 5900f53a1000cf542c51004c
-challengeType: 5
 title: 'Problem 461: Almost Pi'
+challengeType: 5
 forumTopicId: 302136
 ---
 
-## Description
-<section id='description'>
-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|
+# --description--
+
+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|
+
 (where |x| denotes the absolute value of x).
-You are given g(200) = 62 + 752 + 892 + 2262 = 64658.
+
+You are given g(200) = 62 + 752 + 892 + 2262 = 64658.
+
 Find g(10000).
-</section>
 
-## Instructions
-<section id='instructions'>
+# --hints--
 
-</section>
-
-## Tests
-<section id='tests'>
-
-```yml
-tests:
-  - text: <code>euler461()</code> should return 159820276.
-    testString: assert.strictEqual(euler461(), 159820276);
+`euler461()` should return 159820276.
 
+```js
+assert.strictEqual(euler461(), 159820276);
 ```
 
-</section>
+# --seed--
 
-## Challenge Seed
-<section id='challengeSeed'>
-
-<div id='js-seed'>
+## --seed-contents--
 
 ```js
 function euler461() {
@@ -47,17 +42,8 @@ function euler461() {
 euler461();
 ```
 
-</div>
-
-
-
-</section>
-
-## Solution
-<section id='solution'>
+# --solutions--
 
 ```js
 // solution required
 ```
-
-</section>