51 lines
1.5 KiB
Markdown
51 lines
1.5 KiB
Markdown
---
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id: 5900f3db1000cf542c50feee
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title: 'Problem 111: Primes with runs'
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challengeType: 5
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forumTopicId: 301736
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dashedName: problem-111-primes-with-runs
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---
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# --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 M(4, d) N(4, d) S(4, d) 0 2 13 67061 1 3 9 22275 2 3 1 2221 3 3 12 46214 4 3 2 8888 5 3 1 5557 6 3 1 6661 7 3 9 57863 8 3 1 8887 9 3 7 48073
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For d = 0 to 9, the sum of all S(4, d) is 273700. Find the sum of all S(10, d).
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# --hints--
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`euler111()` should return 612407567715.
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```js
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assert.strictEqual(euler111(), 612407567715);
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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 euler111() {
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return true;
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}
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euler111();
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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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