freeCodeCamp/smallest-common-multiple.english.md at 5ea745d61fa4c4b2650c05f28fb1f63f0e8751b6

Mo Zargham 437ba8b103 fix(curriculum): Read-search-ask link now point to correct url as noted in the issue (#37753 )

* fix: broken Read-search-ask link now point to correct url

* fix: changed link to original forum link with more views

* fix: changed http links to correct version

* fix: link in help modal

2019-11-19 19:54:48 -05:00

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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. Remember to use Read-Search-Ask if you get stuck. Try to pair program. Write your own code.

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);
}

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