97 lines
1.6 KiB
Markdown
97 lines
1.6 KiB
Markdown
---
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id: 5a23c84252665b21eecc7ec1
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title: Iterated digits squaring
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challengeType: 5
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forumTopicId: 302291
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dashedName: iterated-digits-squaring
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---
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# --description--
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If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:
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<pre>15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89
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7 -> 49 -> 97 -> 130 -> 10 -> 1
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</pre>
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# --instructions--
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Write a function that takes a number as a parameter and returns 1 or 89 after performing the mentioned process.
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# --hints--
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`iteratedSquare` should be a function.
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```js
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assert(typeof iteratedSquare == 'function');
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```
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`iteratedSquare(4)` should return a number.
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```js
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assert(typeof iteratedSquare(4) == 'number');
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```
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`iteratedSquare(4)` should return `89`.
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```js
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assert.equal(iteratedSquare(4), 89);
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```
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`iteratedSquare(7)` should return `1`.
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```js
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assert.equal(iteratedSquare(7), 1);
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```
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`iteratedSquare(15)` should return `89`.
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```js
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assert.equal(iteratedSquare(15), 89);
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```
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`iteratedSquare(20)` should return `89`.
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```js
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assert.equal(iteratedSquare(20), 89);
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```
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`iteratedSquare(70)` should return `1`.
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```js
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assert.equal(iteratedSquare(70), 1);
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```
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`iteratedSquare(100)` should return `1`.
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```js
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assert.equal(iteratedSquare(100), 1);
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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 iteratedSquare(n) {
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}
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```
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# --solutions--
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```js
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function iteratedSquare(n) {
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var total;
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while (n != 89 && n != 1) {
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total = 0;
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while (n > 0) {
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total += Math.pow(n % 10, 2);
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n = Math.floor(n/10);
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}
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n = total;
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}
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return n;
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}
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```
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