# Problem 111: Primes with runs

![graphic](img111.gif)

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).