ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
85 lines
1.7 KiB
Markdown
85 lines
1.7 KiB
Markdown
---
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id: 5900f3811000cf542c50fe94
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title: 'Problem 21: Amicable numbers'
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challengeType: 5
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forumTopicId: 301851
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dashedName: problem-21-amicable-numbers
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---
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# --description--
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Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).
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If d(`a`) = `b` and d(`b`) = `a`, where `a` ≠ `b`, then `a` and `b` are an amicable pair and each of `a` and `b` 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 `n`.
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# --hints--
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`sumAmicableNum(1000)` should return a number.
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```js
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assert(typeof sumAmicableNum(1000) === 'number');
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```
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`sumAmicableNum(1000)` should return 504.
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```js
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assert.strictEqual(sumAmicableNum(1000), 504);
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```
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`sumAmicableNum(2000)` should return 2898.
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```js
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assert.strictEqual(sumAmicableNum(2000), 2898);
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```
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`sumAmicableNum(5000)` should return 8442.
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```js
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assert.strictEqual(sumAmicableNum(5000), 8442);
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```
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`sumAmicableNum(10000)` should return 31626.
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```js
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assert.strictEqual(sumAmicableNum(10000), 31626);
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```
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# --seed--
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## --seed-contents--
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```js
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function sumAmicableNum(n) {
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return n;
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}
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sumAmicableNum(10000);
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```
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# --solutions--
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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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