Project Euler > Problem 71 > Ordered fractions (Java Solution)

Problem:

Consider the fraction, n/d, where n and d are positive integers. If n[<]d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d [≤] 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d [≤] 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.

Solution:

31626

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 21
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p021 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p021().run());
	}
	
	
	public String run() {
		int sum = 0;
		for (int i = 1; i < 10000; i++) {
			if (isAmicable(i))
				sum += i;
		}
		return Integer.toString(sum);
	}
	
	
	private static boolean isAmicable(int n) {
		int m = divisorSum(n);
		return m != n && divisorSum(m) == n;
	}
	
	
	private static int divisorSum(int n) {
		int sum = 0;
		for (int i = 1; i < n; i++) {
			if (n % i == 0)
				sum += i;
		}
		return sum;
	}
	
}