freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-21-amicable-numbers.md

---
id: 5900f3811000cf542c50fe94
title: 'Problem 21: Amicable numbers'
challengeType: 5
forumTopicId: 301851
dashedName: problem-21-amicable-numbers
---

# --description--

Let d(`n`) be defined as the sum of proper divisors of `n` (numbers less than `n` which divide evenly into `n`).

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.

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.

Evaluate the sum of all the amicable numbers under `n`.

# --hints--

`sumAmicableNum(1000)` should return a number.

```js
assert(typeof sumAmicableNum(1000) === 'number');
```

`sumAmicableNum(1000)` should return 504.

```js
assert.strictEqual(sumAmicableNum(1000), 504);
```

`sumAmicableNum(2000)` should return 2898.

```js
assert.strictEqual(sumAmicableNum(2000), 2898);
```

`sumAmicableNum(5000)` should return 8442.

```js
assert.strictEqual(sumAmicableNum(5000), 8442);
```

`sumAmicableNum(10000)` should return 31626.

```js
assert.strictEqual(sumAmicableNum(10000), 31626);
```

# --seed--

## --seed-contents--

```js
function sumAmicableNum(n) {

  return n;
}

sumAmicableNum(10000);
```

# --solutions--

```js
const sumAmicableNum = (n) => {
  const fsum = (n) => {
    let sum = 1;
    for (let i = 2; i <= Math.floor(Math.sqrt(n)); i++)
      if (Math.floor(n % i) === 0)
        sum += i + Math.floor(n / i);
    return sum;
  };
  let d = [];
  let amicableSum = 0;
  for (let i=2; i<n; i++) d[i] = fsum(i);
  for (let i=2; i<n; i++) {
    let dsum = d[i];
    if (d[dsum]===i && i!==dsum) amicableSum += i+dsum;
  }
  return amicableSum/2;
};
```