## 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).

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;	}	}

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