84 lines
2.3 KiB
Markdown
84 lines
2.3 KiB
Markdown
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---
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id: 5
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localeTitle: 5900f3811000cf542c50fe94
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challengeType: 5
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title: 'Problem 21: Amicable numbers'
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---
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## Description
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<section id='description'>
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Deje que d ( <var>n</var> ) se defina como la suma de los divisores propios de <var>n</var> (números menores que <var>n</var> que se dividen uniformemente en <var>n</var> ).
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Si d ( <var>a</var> ) = <var>b</var> y d ( <var>b</var> ) = <var>a</var> , donde <var>a</var> ≠ <var>b</var> , entonces <var>a</var> y <var>b</var> son un par amigable y cada uno de <var>a</var> y <var>b</var> se llama números amistosos.
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Por ejemplo, los divisores propios de 220 son 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 y 110; por lo tanto, d (220) = 284. Los divisores propios de 284 son 1, 2, 4, 71 y 142; entonces d (284) = 220.
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Evalúa la suma de todos los números amistosos debajo de <var>n</var> .
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>sumAmicableNum(1000)</code> debe devolver 504.
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testString: 'assert.strictEqual(sumAmicableNum(1000), 504, "<code>sumAmicableNum(1000)</code> should return 504.");'
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- text: <code>sumAmicableNum(2000)</code> debe devolver 2898.
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testString: 'assert.strictEqual(sumAmicableNum(2000), 2898, "<code>sumAmicableNum(2000)</code> should return 2898.");'
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- text: <code>sumAmicableNum(5000)</code> debe devolver 8442.
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testString: 'assert.strictEqual(sumAmicableNum(5000), 8442, "<code>sumAmicableNum(5000)</code> should return 8442.");'
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- text: <code>sumAmicableNum(10000)</code> debe devolver 31626.
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testString: 'assert.strictEqual(sumAmicableNum(10000), 31626, "<code>sumAmicableNum(10000)</code> should return 31626.");'
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function sumAmicableNum(n) {
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// Good luck!
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return n;
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}
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sumAmicableNum(10000);
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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const sumAmicableNum = (n) => {
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const fsum = (n) => {
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let sum = 1;
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for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
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if (Math.floor(n % i) === 0)
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sum += i + Math.floor(n / i);
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return sum;
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};
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let d = [];
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let amicableSum = 0;
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for (let i=2; i<n; i++) d[i] = fsum(i);
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for (let i=2; i<n; i++) {
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let dsum = d[i];
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if (d[dsum]===i && i!==dsum) amicableSum += i+dsum;
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}
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return amicableSum/2;
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};
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```
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</section>
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