Project Euler

Problem #33

The fraction ^(49)/_(98) is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that ^(49)/_(98) = ^(4)/_(8), which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, ^(30)/_(50) = ^(3)/_(5), to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

Ruby: Running time = 0.12s
+#Ratio

+#dig_split

def p33
  frax=[]
  (11..99).each do |d|
    (10...d).each do |n|
      nn=dig_split(n)
      dd=dig_split(d)
      nn.each do |dig|
        next if dig==0
        if(dd.include? dig)
          orig=Ratio.new(n,d)
	  newer=Ratio.new(nn[(1+nn.index(dig))%2],dd[(1+dd.index(dig))%2])
	  frax.push orig if orig==newer
	end
      end
    end
  end
  puts frax.product.den
end