32fac23a2d
* fix: clean-up Project Euler 161-180 * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
50 lines
1005 B
Markdown
50 lines
1005 B
Markdown
---
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id: 5900f4181000cf542c50ff2a
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title: >-
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Problem 171: Finding numbers for which the sum of the squares of the digits is
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a square
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challengeType: 5
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forumTopicId: 301806
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dashedName: >-
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problem-171-finding-numbers-for-which-the-sum-of-the-squares-of-the-digits-is-a-square
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---
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# --description--
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For a positive integer $n$, let $f(n)$ be the sum of the squares of the digits (in base 10) of $n$, e.g.
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$$\begin{align}
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& f(3) = 3^2 = 9 \\\\
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& f(25) = 2^2 + 5^2 = 4 + 25 = 29 \\\\
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& f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 \\\\
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\end{align}$$
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Find the last nine digits of the sum of all $n$, $0 < n < {10}^{20}$, such that $f(n)$ is a perfect square.
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# --hints--
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`lastDigitsSumOfPerfectSquare()` should return `142989277`.
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```js
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assert.strictEqual(lastDigitsSumOfPerfectSquare(), 142989277);
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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 lastDigitsSumOfPerfectSquare() {
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return true;
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}
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lastDigitsSumOfPerfectSquare();
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```
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# --solutions--
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```js
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// solution required
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```
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