Project Euler > Problem 130 > Composites with prime repunit property (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(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.

You are given that for all primes, p [>] 5, that p [−] 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.

However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.

Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n [−] 1 is divisible by A(n).

Solution:

443839

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 30
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p030 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p030().run());
	}
	
	
	public String run() {
		// As stated in the problem, 1 = 1^5 is excluded.
		// If a number has at least n >= 7 digits, then even if every digit is 9,
		// n * 9^5 is still less than the number (which is at least 10^n).
		int sum = 0;
		for (int i = 2; i < 1000000; i++) {
			if (i == fifthPowerDigitSum(i))
				sum += i;
		}
		return Integer.toString(sum);
	}
	
	
	private static int fifthPowerDigitSum(int x) {
		int sum = 0;
		while (x != 0) {
			int y = x % 10;
			sum += y * y * y * y * y;
			x /= 10;
		}
		return sum;
	}
	
}