## Problem:

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:

1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:

169 [→] 363601 [→] 1454 [→] 169
871 [→] 45361 [→] 871
872 [→] 45362 [→] 872

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

69 [→] 363600 [→] 1454 [→] 169 [→] 363601 ([→] 1454)
78 [→] 45360 [→] 871 [→] 45361 ([→] 871)
540 [→] 145 ([→] 145)

Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

2783915460

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 24 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p024 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p024().run());	}			public String run() {		// Initialize		int[] array = new int;		for (int i = 0; i < array.length; i++)			array[i] = i;				// Permute		for (int i = 0; i < 999999; i++) {			if (!Library.nextPermutation(array))				throw new AssertionError();		}				// Format output		String ans = "";		for (int i = 0; i < array.length; i++)			ans += array[i];		return ans;	}	}