Problem 57

Projekt Euler w Ruby
1

Problem 57

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?

2

[code=ruby]#!/usr/bin/env ruby
N = 1000

def sqrt2(i)
n, d = 1, 2
(i-1).times { n, d = d, 2*d+n }
n += d
[n, d]
end

x = 0
(1…N).each do |i|
n, d = sqrt2(i)
x += 1 if n.to_s.length > d.to_s.length
end
p x[/code]

3

[code=ruby]p (0…1000).select {|i|
l,m = 1,2
i.times {
l += (m*2)
l,m = m,l
}
l += m
l.to_s.size > m.to_s.size

}.size[/code]