2.4 KiB
2.4 KiB
title | id | challengeType |
---|---|---|
Abundant, deficient and perfect number classifications | 594810f028c0303b75339acd | 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
tests:
- text: <code>getDPA</code> is a function.
testString: assert(typeof getDPA === 'function', '<code>getDPA</code> is a function.');
- text: <code>getDPA</code> should return an array.
testString: assert(Array.isArray(getDPA(100)), '<code>getDPA</code> should return an array.');
- text: <code>getDPA</code> return value should have a length of 3.
testString: assert(getDPA(100).length === 3, '<code>getDPA</code> return value should have a length of 3.');
- text: <code>getDPA(20000)</code> should equal [15043, 4, 4953]
testString: assert.deepEqual(getDPA(20000), solution, '<code>getDPA(20000)</code> should equal [15043, 4, 4953]');
Challenge Seed
function getDPA (num) {
// Good luck!
}
After Test
const solution = [15043, 4, 4953];
Solution
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;
}