--- title: 'Abundant, deficient and perfect number classifications' id: 594810f028c0303b75339acd challengeType: 5 --- ## Description

These define three classifications of positive integers based on their proper divisors.

Let $P(n)$ be the sum of the proper divisors of n where proper divisors are all positive integers n other than n itself.

If P(n) < n then n is classed as "deficient"

If P(n) === n then n is classed as "perfect"

If P(n) > n then n is classed as "abundant"

Example:

6 has proper divisors of 1, 2, and 3.

1 + 2 + 3 = 6, so 6 is classed as a perfect number.

Implement a function that calculates how many of the integers from 1 to 20,000 (inclusive) are in each of the three classes. Output the result as an array in the following format [deficient, perfect, abundant].

## Instructions
## Tests
```yml tests: - text: getDPA is a function. testString: assert(typeof getDPA === 'function', 'getDPA is a function.'); - text: getDPA should return an array. testString: assert(Array.isArray(getDPA(100)), 'getDPA should return an array.'); - text: getDPA return value should have a length of 3. testString: assert(getDPA(100).length === 3, 'getDPA return value should have a length of 3.'); - text: getDPA(20000) should equal [15043, 4, 4953] testString: assert.deepEqual(getDPA(20000), solution, 'getDPA(20000) should equal [15043, 4, 4953]'); ```
## Challenge Seed
```js function getDPA (num) { // Good luck! } ```
### After Test
```js const solution = [15043, 4, 4953]; ```
## Solution
```js function getDPA (num) { const dpa = [1, 0, 0]; for (let n = 2; n <= num; n += 1) { let ds = 1; const e = Math.sqrt(n); for (let d = 2; d < e; d += 1) { if (n % d === 0) { ds += d + (n / d); } } if (n % e === 0) { ds += e; } dpa[ds < n ? 0 : ds === n ? 1 : 2] += 1; } return dpa; } ```