Eccoci.

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.

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

Find the sum of all $S(10, d)$.