freeCodeCamp/curriculum/challenges/english/02-javascript-algorithms-and-data-structures/intermediate-algorithm-scripting/smallest-common-multiple.english.md

id, title, isRequired, challengeType, forumTopicId

id	title	isRequired	challengeType	forumTopicId
ae9defd7acaf69703ab432ea	Smallest Common Multiple	true	5	16075

Description

Find the smallest common multiple of the provided parameters that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters. The range will be an array of two numbers that will not necessarily be in numerical order. For example, if given 1 and 3, find the smallest common multiple of both 1 and 3 that is also evenly divisible by all numbers between 1 and 3. The answer here would be 6.

Instructions

Tests

tests:
  - text: <code>smallestCommons([1, 5])</code> should return a number.
    testString: assert.deepEqual(typeof smallestCommons([1, 5]), 'number');
  - text: <code>smallestCommons([1, 5])</code> should return 60.
    testString: assert.deepEqual(smallestCommons([1, 5]), 60);
  - text: <code>smallestCommons([5, 1])</code> should return 60.
    testString: assert.deepEqual(smallestCommons([5, 1]), 60);
  - text: <code>smallestCommons([2, 10])</code> should return 2520.
    testString: assert.deepEqual(smallestCommons([2, 10]), 2520);
  - text: <code>smallestCommons([1, 13])</code> should return 360360.
    testString: assert.deepEqual(smallestCommons([1, 13]), 360360);
  - text: <code>smallestCommons([23, 18])</code> should return 6056820.
    testString: assert.deepEqual(smallestCommons([23, 18]), 6056820);

Challenge Seed

function smallestCommons(arr) {
  return arr;
}


smallestCommons([1,5]);

Solution

function gcd(a, b) {
    while (b !== 0) {
        a = [b, b = a % b][0];
    }
    return a;
}

function lcm(a, b) {
    return (a * b) / gcd(a, b);
}

function smallestCommons(arr) {
  arr.sort(function(a,b) {return a-b;});
  var rng = [];
  for (var i = arr[0]; i <= arr[1]; i++) {
    rng.push(i);
  }
  return rng.reduce(lcm);
}