62 lines
2.0 KiB
Markdown
62 lines
2.0 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 | $M(4, d)$ | $N(4, d)$ | $S(4, d)$ |
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| -------- | --------- | --------- | --------- |
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| 0 | 2 | 13 | 67061 |
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| 1 | 3 | 9 | 22275 |
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| 2 | 3 | 1 | 2221 |
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| 3 | 3 | 12 | 46214 |
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| 4 | 3 | 2 | 8888 |
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| 5 | 3 | 1 | 5557 |
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| 6 | 3 | 1 | 6661 |
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| 7 | 3 | 9 | 57863 |
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| 8 | 3 | 1 | 8887 |
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| 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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`primesWithRuns()` should return `612407567715`.
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```js
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assert.strictEqual(primesWithRuns(), 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 primesWithRuns() {
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return true;
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}
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primesWithRuns();
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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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