Project Euler > Problem 108 > Diophantine reciprocals I (Java Solution)

Problem:

In the following equation x, y, and n are positive integers.

1

x
+
1

y
=
1

n

For n = 4 there are exactly three distinct solutions:

1

5
+
1

20
=
1

4
1

6
+
1

12
=
1

4
1

8
+
1

8
=
1

4

What is the least value of n for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of problem 110; it is strongly advised that you solve this one first.

Solution:

40824

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 8
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p008 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p008().run());
	}
	
	
	private static final String NUMBER = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";
	
	
	public String run() {
		int maxProd = -1;
		for (int i = 0; i + 5 <= NUMBER.length(); i++) {
			int prod = 1;
			for (int j = 0; j < 5; j++)
				prod *= NUMBER.charAt(i + j) - '0';
			maxProd = Math.max(prod, maxProd);
		}
		return Integer.toString(maxProd);
	}
	
}