Project Euler Lab - Problem 111

#111 - Primes with Runs

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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 \(S(10, d)\).

This problem is taken from Project Euler, Problem 111.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=111. Published Friday, 16th December 2005, 06:00 pm. Solved by 8,572 members at time of mirroring.

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