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

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+---
+id: 5900f3ad1000cf542c50fec0
+title: 'Problem 65: Convergents of e'
+challengeType: 5
+forumTopicId: 302177
+dashedName: problem-65-convergents-of-e
+---
+
+# --description--
+
+The square root of 2 can be written as an infinite continued fraction.
+
+$\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}$
+
+The infinite continued fraction can be written, $\\sqrt{2} = \[1; (2)]$ indicates that 2 repeats *ad infinitum*. In a similar way, $\\sqrt{23} = \[4; (1, 3, 1, 8)]$. It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\\sqrt{2}$.
+
+$1 + \\dfrac{1}{2} = \\dfrac{3}{2}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2}} = \\dfrac{7}{5}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}} = \\dfrac{17}{12}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}}} = \\dfrac{41}{29}$
+
+Hence the sequence of the first ten convergents for $\\sqrt{2}$ are:
+
+$1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{3363}{2378}, ...$
+
+What is most surprising is that the important mathematical constant, $e = \[2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$. The first ten terms in the sequence of convergents for `e` are:
+
+$2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...$
+
+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 `n`<sup>th</sup> convergent of the continued fraction for `e`.
+
+# --hints--
+
+`convergentsOfE(10)` should return a number.
+
+```js
+assert(typeof convergentsOfE(10) === 'number');
+```
+
+`convergentsOfE(10)` should return `17`.
+
+```js
+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--
+
+## --seed-contents--
+
+```js
+function convergentsOfE(n) {
+
+  return true;
+}
+
+convergentsOfE(10);
+```
+
+# --solutions--
+
+```js
+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]);
+}
+```