91 lines
1.7 KiB
Markdown
91 lines
1.7 KiB
Markdown
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---
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id: ae9defd7acaf69703ab432ea
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title: Smallest Common Multiple
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challengeType: 5
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forumTopicId: 16075
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dashedName: smallest-common-multiple
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---
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# --description--
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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.
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The range will be an array of two numbers that will not necessarily be in numerical order.
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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.
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# --hints--
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`smallestCommons([1, 5])` should return a number.
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```js
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assert.deepEqual(typeof smallestCommons([1, 5]), 'number');
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```
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`smallestCommons([1, 5])` should return 60.
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```js
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assert.deepEqual(smallestCommons([1, 5]), 60);
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```
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`smallestCommons([5, 1])` should return 60.
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```js
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assert.deepEqual(smallestCommons([5, 1]), 60);
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```
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`smallestCommons([2, 10])` should return 2520.
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```js
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assert.deepEqual(smallestCommons([2, 10]), 2520);
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```
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`smallestCommons([1, 13])` should return 360360.
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```js
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assert.deepEqual(smallestCommons([1, 13]), 360360);
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```
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`smallestCommons([23, 18])` should return 6056820.
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```js
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assert.deepEqual(smallestCommons([23, 18]), 6056820);
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```
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# --seed--
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## --seed-contents--
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```js
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function smallestCommons(arr) {
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return arr;
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}
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smallestCommons([1,5]);
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```
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# --solutions--
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```js
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function gcd(a, b) {
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while (b !== 0) {
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a = [b, b = a % b][0];
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}
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return a;
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}
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function lcm(a, b) {
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return (a * b) / gcd(a, b);
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}
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function smallestCommons(arr) {
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arr.sort(function(a,b) {return a-b;});
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var rng = [];
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for (var i = arr[0]; i <= arr[1]; i++) {
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rng.push(i);
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}
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return rng.reduce(lcm);
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}
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```
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