freeCodeCamp/curriculum/challenges/italian/10-coding-interview-prep/rosetta-code/least-common-multiple.md

88 lines
2.0 KiB
Markdown
Raw Normal View History

---
id: 5a23c84252665b21eecc7edf
title: Least common multiple
challengeType: 5
forumTopicId: 302301
dashedName: least-common-multiple
---
# --description--
2021-07-10 04:23:54 +00:00
The least common multiple of 12 and 18 is 36, because 12 is a factor (12 × 3 = 36), and 18 is a factor (18 × 2 = 36), and there is no positive integer less than 36 that has both factors. As a special case, if either *m* or *n* is zero, then the least common multiple is zero. One way to calculate the least common multiple is to iterate all the multiples of *m*, until you find one that is also a multiple of *n*. If you already have *gcd* for [greatest common divisor](https://rosettacode.org/wiki/greatest common divisor), then this formula calculates *lcm*. ( \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)} )
# --instructions--
Compute the least common multiple of an array of integers. Given *m* and *n*, the least common multiple is the smallest positive integer that has both *m* and *n* as factors.
# --hints--
`LCM` should be a function.
```js
assert(typeof LCM == 'function');
```
`LCM([2, 4, 8])` should return a number.
```js
assert(typeof LCM([2, 4, 8]) == 'number');
```
`LCM([2, 4, 8])` should return `8`.
```js
assert.equal(LCM([2, 4, 8]), 8);
```
`LCM([4, 8, 12])` should return `24`.
```js
assert.equal(LCM([4, 8, 12]), 24);
```
`LCM([3, 4, 5, 12, 40])` should return `120`.
```js
assert.equal(LCM([3, 4, 5, 12, 40]), 120);
```
`LCM([11, 33, 90])` should return `990`.
```js
assert.equal(LCM([11, 33, 90]), 990);
```
`LCM([-50, 25, -45, -18, 90, 447])` should return `67050`.
```js
assert.equal(LCM([-50, 25, -45, -18, 90, 447]), 67050);
```
# --seed--
## --seed-contents--
```js
function LCM(A) {
}
```
# --solutions--
```js
function LCM(A) {
var n = A.length,
a = Math.abs(A[0]);
for (var i = 1; i < n; i++) {
var b = Math.abs(A[i]),
c = a;
while (a && b) {
a > b ? (a %= b) : (b %= a);
}
a = Math.abs(c * A[i]) / (a + b);
}
return a;
}
```