yaworsw/euler-manager

---
:id: 111
:name: Primes with runs
:url: https://projecteuler.net/problem=111
:content: |+
  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_).