chore(i8n,learn): processed translations

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

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+---
+id: 5a23c84252665b21eecc7edf
+title: Least common multiple
+challengeType: 5
+forumTopicId: 302301
+dashedName: least-common-multiple
+---
+
+# --description--
+
+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;
+}
+```