Problem:

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).

70600674

Code:The solution may include methods that will be found here: Library.java .

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 11 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p011 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p011().run());	}			public String run() {		int max = -1;		max = Math.max(maxProduct(1, 0), max);		max = Math.max(maxProduct(0, 1), max);		max = Math.max(maxProduct(1, 1), max);		max = Math.max(maxProduct(1, -1), max);		return Integer.toString(max);	}			private static int maxProduct(int dx, int dy) {		int max = -1;		for (int y = 0; y < SQUARE.length; y++) {			for (int x = 0; x < SQUARE[y].length; x++)				max = Math.max(product(x, y, dx, dy, 4), max);		}		return max;	}			private static int product(int x, int y, int dx, int dy, int n) {		// First endpoint is assumed to be in bounds. Check if second endpoint is in bounds.		if (!isInBounds(x + (n - 1) * dx, y + (n - 1) * dy))			return -1;				int prod = 1;		for (int i = 0; i < n; i++, x += dx, y += dy)			prod *= SQUARE[y][x];		return prod;	}			private static boolean isInBounds(int x, int y) {		return 0 <= y && y < SQUARE.length && 0 <= x && x < SQUARE[y].length;	}			private static int[][] SQUARE = {		{ 8,02,22,97,38,15,00,40,00,75,04,05,07,78,52,12,50,77,91, 8},		{49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,04,56,62,00},		{81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,03,49,13,36,65},		{52,70,95,23,04,60,11,42,69,24,68,56,01,32,56,71,37,02,36,91},		{22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80},		{24,47,32,60,99,03,45,02,44,75,33,53,78,36,84,20,35,17,12,50},		{32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70},		{67,26,20,68,02,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21},		{24,55,58,05,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72},		{21,36,23, 9,75,00,76,44,20,45,35,14,00,61,33,97,34,31,33,95},		{78,17,53,28,22,75,31,67,15,94,03,80,04,62,16,14, 9,53,56,92},		{16,39,05,42,96,35,31,47,55,58,88,24,00,17,54,24,36,29,85,57},		{86,56,00,48,35,71,89,07,05,44,44,37,44,60,21,58,51,54,17,58},		{19,80,81,68,05,94,47,69,28,73,92,13,86,52,17,77,04,89,55,40},		{04,52, 8,83,97,35,99,16,07,97,57,32,16,26,26,79,33,27,98,66},		{88,36,68,87,57,62,20,72,03,46,33,67,46,55,12,32,63,93,53,69},		{04,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36},		{20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,04,36,16},		{20,73,35,29,78,31,90,01,74,31,49,71,48,86,81,16,23,57,05,54},		{01,70,54,71,83,51,54,69,16,92,33,48,61,43,52,01,89,19,67,48}	};	}