fix(curriculum): rework Problem Euler 65 (#41790)

* fix: rework challenge to use argument in function

* fix: add solution

* fix: missing backticks

curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md

@@ -26,20 +26,44 @@ $2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{1
 
 The sum of digits in the numerator of the 10<sup>th</sup> convergent is $1 + 4 + 5 + 7 = 17$.
 
-Find the sum of digits in the numerator of the 100<sup>th</sup> convergent of the continued fraction for `e`.
+Find the sum of digits in the numerator of the `n`<sup>th</sup> convergent of the continued fraction for `e`.
 
 # --hints--
 
-`convergentsOfE()` should return a number.
+`convergentsOfE(10)` should return a number.
 
 ```js
-assert(typeof convergentsOfE() === 'number');
+assert(typeof convergentsOfE(10) === 'number');
 ```
 
-`convergentsOfE()` should return 272.
+`convergentsOfE(10)` should return `17`.
 
 ```js
-assert.strictEqual(convergentsOfE(), 272);
+assert.strictEqual(convergentsOfE(10), 17);
+```
+
+`convergentsOfE(30)` should return `53`.
+
+```js
+assert.strictEqual(convergentsOfE(30), 53);
+```
+
+`convergentsOfE(50)` should return `91`.
+
+```js
+assert.strictEqual(convergentsOfE(50), 91);
+```
+
+`convergentsOfE(70)` should return `169`.
+
+```js
+assert.strictEqual(convergentsOfE(70), 169);
+```
+
+`convergentsOfE(100)` should return `272`.
+
+```js
+assert.strictEqual(convergentsOfE(100), 272);
 ```
 
 # --seed--
@@ -47,16 +71,47 @@ assert.strictEqual(convergentsOfE(), 272);
 ## --seed-contents--
 
 ```js
-function convergentsOfE() {
+function convergentsOfE(n) {
 
   return true;
 }
 
-convergentsOfE();
+convergentsOfE(10);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function convergentsOfE(n) {
+  function sumDigits(num) {
+    let sum = 0n;
+    while (num > 0) {
+      sum += num % 10n;
+      num = num / 10n;
+    }
+    return parseInt(sum);
+  }
+
+  // BigInt is needed for high convergents
+  let convergents = [
+    [2n, 1n],
+    [3n, 1n]
+  ];
+  const multipliers = [1n, 1n, 2n];
+  for (let i = 2; i < n; i++) {
+    const [secondLastConvergent, lastConvergent] = convergents;
+    const [secondLastNumerator, secondLastDenominator] = secondLastConvergent;
+    const [lastNumerator, lastDenominator] = lastConvergent;
+    const curMultiplier = multipliers[i % 3];
+
+    const numerator = secondLastNumerator + curMultiplier * lastNumerator;
+    const denominator = secondLastDenominator + curMultiplier * lastDenominator;
+
+    convergents = [lastConvergent, [numerator, denominator]]
+    if (i % 3 === 2) {
+      multipliers[2] += 2n;
+    }
+  }
+  return sumDigits(convergents[1][0]);
+}
 ```