Question
The fraction is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that , which is correct, is obtained by cancelling the 9s.
We shall consider fractions like, , 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.
Haskell
import Data.Ratio (denominator)
fraction :: Int -> Int -> Rational
fraction n d = fromIntegral n / fromIntegral d
curious :: Int -> Int -> Bool
curious n d | f > 1 = False
| d1 == 0 || d2 == 0 = False
| n == d = False
| n2 /= d1 = False
| otherwise = fraction n1 d2 == f
where f = fraction n d
(n1, n2) = quotRem n 10
(d1, d2) = quotRem d 10
main :: IO ()
main = print $ denominator $ product [fraction n d | n <- [10..99], d <- [10..99], curious n d]
$ ghc -O2 -o curious-fractions curious-fractions.hs
$ time ./curious-fractions
real 0m0.003s
user 0m0.000s
sys 0m0.003s
Python
#!/usr/bin/env python
from fractions import *
from itertools import *
from functools import reduce
def is_curious(n, d):
f = Fraction(n, d)
if f >= 1:
return False
n_digits = [int(digit) for digit in str(n)]
d_digits = [int(digit) for digit in str(d)]
if n_digits[1] == d_digits[0]:
try:
if Fraction(n_digits[0], d_digits[1]) == f:
return True
except:
pass
return False
def main():
fractions = product(list(range(10, 100)), list(range(10, 100)))
print(reduce(lambda a, b: a * b, (Fraction(*f) for f in fractions if is_curious(*f))).denominator)
if __name__ == "__main__":
main()
Ruby
#!/usr/bin/env ruby
puts ('10'..'99').to_a.product(('10'..'99').to_a).select { |num, den|
(num[0].to_f / den[1].to_f) == (num.to_f / den.to_f) && num[1] == den[0] && num[1] != den[1]
}.uniq.map { |num, den| [num.to_i, den.to_i] }.reduce([1, 1]) { |p, v|
f = [p[0] * v[0], p[1] * v[1]]
[f[0] / f[0].gcd(f[1]), f[1] / f[0].gcd(f[1])]
}[1]