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and change all the challenges to new `md` format.
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Oliver Eyton-Williams
2020-11-27 19:02:05 +01:00
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curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-65-convergents-of-e.md
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@@ -1,63 +1,49 @@
---
id: 5900f3ad1000cf542c50fec0
challengeType: 5
title: 'Problem 65: Convergents of e'
challengeType: 5
forumTopicId: 302177
---
## Description
<section id='description'>
# --description--
The square root of 2 can be written as an infinite continued fraction.
$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}}$
$\\sqrt{2} = 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + ...}}}}$
The infinite continued fraction can be written, $\sqrt{2} = [1; (2)]$ indicates that 2 repeats <i>ad infinitum</i>. 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 $\sqrt{2}$.
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 $\\sqrt{2}$.
$1 + \dfrac{1}{2} = \dfrac{3}{2}\\\\
1 + \dfrac{1}{2 + \dfrac{1}{2}} = \dfrac{7}{5}\\\\
1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} = \dfrac{17}{12}\\\\
1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} = \dfrac{41}{29}$
$1 + \\dfrac{1}{2} = \\dfrac{3}{2}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2}} = \\dfrac{7}{5}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}} = \\dfrac{17}{12}\\\\ 1 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2 + \\dfrac{1}{2}}}} = \\dfrac{41}{29}$
Hence the sequence of the first ten convergents for $\sqrt{2}$ are:
Hence the sequence of the first ten convergents for $\\sqrt{2}$ are:
$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ...$
$1, \\dfrac{3}{2}, \\dfrac{7}{5}, \\dfrac{17}{12}, \\dfrac{41}{29}, \\dfrac{99}{70}, \\dfrac{239}{169}, \\dfrac{577}{408}, \\dfrac{1393}{985}, \\dfrac{3363}{2378}, ...$
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 <var>e</var> are:
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:
$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ...$
$2, 3, \\dfrac{8}{3}, \\dfrac{11}{4}, \\dfrac{19}{7}, \\dfrac{87}{32}, \\dfrac{106}{39}, \\dfrac{193}{71}, \\dfrac{1264}{465}, \\dfrac{1457}{536}, ...$
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 100<sup>th</sup> convergent of the continued fraction for <var>e</var>.
Find the sum of digits in the numerator of the 100<sup>th</sup> convergent of the continued fraction for `e`.
</section>
# --hints--
## Instructions
<section id='instructions'>
</section>
## Tests
<section id='tests'>
```yml
tests:
- text: <code>convergentsOfE()</code> should return a number.
testString: assert(typeof convergentsOfE() === 'number');
- text: <code>convergentsOfE()</code> should return 272.
testString: assert.strictEqual(convergentsOfE(), 272);
`convergentsOfE()` should return a number.
```js
assert(typeof convergentsOfE() === 'number');
```
</section>
`convergentsOfE()` should return 272.
## Challenge Seed
<section id='challengeSeed'>
```js
assert.strictEqual(convergentsOfE(), 272);
```
<div id='js-seed'>
# --seed--
## --seed-contents--
```js
function convergentsOfE() {
@@ -68,17 +54,8 @@ function convergentsOfE() {
convergentsOfE();
```
</div>
</section>
## Solution
<section id='solution'>
# --solutions--
```js
// solution required
```
</section>