## Primes with runs

### Problem 111

Published on Friday, 16th December 2005, 06:00 pm; Solved by 3437

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).