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.
[√]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.
Solution:
137846528820
Code:
The solution may include methods that will be found here: Library.java .
1 | public interface EulerSolution{ |
2 |
3 | public String run(); |
4 |
5 | } |
01 | /* |
02 | * Solution to Project Euler problem 15 |
03 | * By Nayuki Minase |
04 | * |
07 | */ |
08 |
09 |
10 | public final class p015 implements EulerSolution { |
11 | |
12 | public static void main(String[] args) { |
13 | System.out.println(new p015().run()); |
14 | } |
15 | |
16 | |
17 | public String run() { |
18 | return Library.binomial(40, 20).toString(); |
19 | } |
20 | |
21 | } |
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