## Problem:

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

[√] 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

104743

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

public interface EulerSolution{

public String run();

}
/**
 * Solution to Project Euler problem 7
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/

public final class p007 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p007().run());
}

public String run() {
for (int i = 2, count = 0; ; i++) {
if (Library.isPrime(i)) {
count++;
if (count == 10001)
return Integer.toString(i);
}
}
}

}