79d9012432
* fix(curriculum): tests quotes * fix(curriculum): fill seed-teardown * fix(curriculum): fix tests and remove unneeded seed-teardown
83 lines
2.3 KiB
Markdown
83 lines
2.3 KiB
Markdown
---
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id: 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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Let d(<var>n</var>) be defined as the sum of proper divisors of <var>n</var> (numbers less than <var>n</var> which divide evenly into <var>n</var>).
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If d(<var>a</var>) = <var>b</var> and d(<var>b</var>) = <var>a</var>, where <var>a</var> ≠ <var>b</var>, then <var>a</var> and <var>b</var> are an amicable pair and each of <var>a</var> and <var>b</var> are called amicable numbers.
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For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
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Evaluate the sum of all the amicable numbers under <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> should return 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> should return 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> should return 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> should return 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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