45 lines
919 B
Markdown
45 lines
919 B
Markdown
---
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id: 5900f5471000cf542c510059
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title: 'Problem 474: Last digits of divisors'
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challengeType: 5
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forumTopicId: 302151
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dashedName: problem-474-last-digits-of-divisors
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---
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# --description--
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For a positive integer $n$ and digits $d$, we define $F(n, d)$ as the number of the divisors of $n$ whose last digits equal $d$.
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For example, $F(84, 4) = 3$. Among the divisors of 84 (1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84), three of them (4, 14, 84) have the last digit 4.
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We can also verify that $F(12!, 12) = 11$ and $F(50!, 123) = 17\\,888$.
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Find $F({10}^6!, 65\\,432) \text{ modulo } ({10}^{16} + 61)$.
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# --hints--
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`lastDigitsOfDivisors()` should return `9690646731515010`.
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```js
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assert.strictEqual(lastDigitsOfDivisors(), 9690646731515010);
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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 lastDigitsOfDivisors() {
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return true;
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}
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lastDigitsOfDivisors();
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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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