22afc2a0ca
Co-authored-by: Oliver Eyton-Williams <ojeytonwilliams@gmail.com> Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> Co-authored-by: Beau Carnes <beaucarnes@gmail.com>
111 lines
1.7 KiB
Markdown
111 lines
1.7 KiB
Markdown
---
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id: 5900f3db1000cf542c50feee
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challengeType: 5
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isHidden: false
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title: 'Problem 111: Primes with runs'
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forumTopicId: 301736
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---
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## Description
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<section id='description'>
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Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:
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1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111
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We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.
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So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.
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In the same way we obtain the following results for 4-digit primes.
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Digit, d
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M(4, d)
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N(4, d)
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S(4, d)
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0
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2
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13
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67061
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1
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3
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9
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22275
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2
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3
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1
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2221
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3
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3
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12
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46214
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4
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3
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2
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8888
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5
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3
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1
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5557
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6
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3
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1
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6661
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7
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3
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9
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57863
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8
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3
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1
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8887
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9
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3
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7
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48073
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For d = 0 to 9, the sum of all S(4, d) is 273700.
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Find the sum of all S(10, d).
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</section>
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## Instructions
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<section id='instructions'>
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: <code>euler111()</code> should return 612407567715.
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testString: assert.strictEqual(euler111(), 612407567715);
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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function euler111() {
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// Good luck!
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return true;
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}
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euler111();
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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</section>
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