Project Euler > Problem 110 > Diophantine reciprocals II (Java Solution)

Problem:

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

1

x
+
1

y
=
1

n

It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for which the total number of distinct solutions exceeds one hundred.

What is the least value of n for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.


Solution:

142913828922

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

public interface EulerSolution{

public String run();

}
/* 
* Solution to Project Euler problem 10
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/


public final class p010 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p010().run());
}


private static final int LIMIT = 2000000;

public String run() {
long sum = 0;
for (int p : Library.listPrimes(LIMIT - 1))
sum += p;
return Long.toString(sum);
}

}


No comments :

Post a Comment

Follow Me

If you like our content, feel free to follow me to stay updated.

Subscribe

Enter your email address:

We hate spam as much as you do.

Upload Material

Got an exam, project, tutorial video, exercise, solutions, unsolved problem, question, solution manual? We are open to any coding material. Why not upload?

Upload

Copyright © 2012 - 2014 Java Problems  --  About  --  Attribution  --  Privacy Policy  --  Terms of Use  --  Contact