## 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 there are 3 fractions between 1/3 and 1/2.

How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d [≤] 12,000?

4179871

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 23 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p023 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p023().run());	}			private static final int LIMIT = 28123;		private boolean[] isAbundant = new boolean[LIMIT + 1];		public String run() {		// Compute look-up table		for (int i = 1; i < isAbundant.length; i++)			isAbundant[i] = isAbundant(i);				int sum = 0;		for (int i = 1; i <= LIMIT; i++) {			if (!isSumOf2Abundants(i))				sum += i;		}		return Integer.toString(sum);	}			private boolean isSumOf2Abundants(int n) {		for (int i = 0; i <= n; i++) {			if (isAbundant[i] && isAbundant[n - i])				return true;		}		return false;	}			private static boolean isAbundant(int n) {		if (n < 1)			throw new IllegalArgumentException();				int sum = 1;  // Sum of factors less than n		int end = Library.sqrt(n);		for (int i = 2; i <= end; i++) {			if (n % i == 0)				sum += i + n / i;		}		if (end * end == n)			sum -= end;		return sum > n;	}	}