## Problem:

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.

For example, R(10) = 1111111111 = 11[×]41[×]271[×]9091, and the sum of these prime factors is 9414.

Find the sum of the first forty prime factors of R(109).

45228

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 32 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */import java.util.Arrays;public final class p032 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p032().run());	}			public String run() {		// A candidate product has at most 4 digits. This is because if it has 5 digits,		// then the two multiplicands must have at least 5 digits put together.		int sum = 0;		for (int i = 1; i < 10000; i++) {			if (hasPandigitalProduct(i))				sum += i;		}		return Integer.toString(sum);	}			private static boolean hasPandigitalProduct(int n) {		// Find and examine all factors of n		for (int i = 1; i <= n; i++) {			if (n % i == 0 && isPandigital("" + n + i + n/i))				return true;		}		return false;	}			private static boolean isPandigital(String s) {		if (s.length() != 9)			return false;		char[] temp = s.toCharArray();		Arrays.sort(temp);		return new String(temp).equals("123456789");	}	}