Project Euler > Problem 132 > Large repunit factors (Java Solution)

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).

Solution:

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");
	}
	
}