freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-111-primes-with-runs.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3db1000cf542c50feee	Problem 111: Primes with runs	5	301736	problem-111-primes-with-runs

--description--

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:

1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111

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.

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.

In the same way we obtain the following results for 4-digit primes.

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

For d = 0 to 9, the sum of all S(4, d) is 273700. Find the sum of all S(10, d).

--hints--

primesWithRuns() should return 612407567715.

assert.strictEqual(primesWithRuns(), 612407567715);

--seed--

--seed-contents--

function primesWithRuns() {

  return true;
}

primesWithRuns();

--solutions--

// solution required