## Problem:

The square root of 2 can be written as an infinite continued fraction.

[√]2 = 1 +
1

2 +
1

2 +
1

2 +
1

2 + ...

The infinite continued fraction can be written, [√]2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, [√]23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for [√]2.

1 +
1

= 3/2

2

1 +
1

= 7/5
2 +
1

2

1 +
1

= 17/12
2 +
1

2 +
1

2

1 +
1

= 41/29
2 +
1

2 +
1

2 +
1

2

Hence the sequence of the first ten convergents for [√]2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

137846528820

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

public interface EulerSolution{public String run();}
/*  * Solution to Project Euler problem 15 * By Nayuki Minase *  * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */public final class p015 implements EulerSolution {		public static void main(String[] args) {		System.out.println(new p015().run());	}			public String run() {		return Library.binomial(40, 20).toString();	}	}