2.0 KiB
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 594810f028c0303b75339acd | Abundant, deficient and perfect number classifications | 5 | 302221 | abundant-deficient-and-perfect-number-classifications |
--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.
--instructions--
Implement a function that calculates how many of the integers from 1 to num (inclusive) are in each of the three classes. Output the result as an array in the following format [deficient, perfect, abundant].
--hints--
getDPA should be a function.
assert(typeof getDPA === 'function');
getDPA(5000) should return an array.
assert(Array.isArray(getDPA(5000)));
getDPA(5000) return array should have a length of 3.
assert(getDPA(5000).length === 3);
getDPA(5000) should return [3758, 3, 1239].
assert.deepEqual(getDPA(5000), [3758, 3, 1239]);
getDPA(10000) should return [7508, 4, 2488].
assert.deepEqual(getDPA(10000), [7508, 4, 2488]);
getDPA(20000) should return [15043, 4, 4953].
assert.deepEqual(getDPA(20000), [15043, 4, 4953]);
--seed--
--seed-contents--
function getDPA(num) {
}
--solutions--
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;
}