﻿# -*- coding: utf-8 -*-
# ProjectEuler/src/python/problem111.py
#
# Primes with runs
# ================
# Published on Friday, 16th December 2005, 06:00 pm
#
# 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).

import projecteuler as pe

def main():
    pass

if __name__ == "__main__":
    main()
