## 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.

The least value of n for which A(n) first exceeds ten is 17.

Find the least value of n for which A(n) first exceeds one-million.

9183

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 29 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */import java.math.BigInteger;import java.util.HashSet;import java.util.Set;public final class p029 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p029().run());	}			public String run() {		Set<BigInteger> generated = new HashSet<BigInteger>();		for (int a = 2; a <= 100; a++) {			for (int b = 2; b <= 100; b++)				generated.add(BigInteger.valueOf(a).pow(b));		}		return Integer.toString(generated.size());	}	}