Problem 33 - Digit Cancelling Fractions

Problem

This problem comes from Project Euler 33

Problem

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.

Solution

I took the most straightforward solution I could think of (cause I’m a straightforward kind of guy)

1. Iterate through a list of two-digit numerators (n) and denominators (d), where the d > n (so that n/d < 1).
2. Ensure that we skip trivial examples. That is, where n and d are divisible by 10.
3. Take any digits that matches in both the numerator and denominator and remove them. If the fraction is still the same then we found a number.
4. Take the product of all such numbers and use the denominator as the answer.

Code

# Project Euler: Problem 33
# Digit Cancelling Fractions

from fractions import Fraction

prod = 1
for n in range(11, 100):
    for d in range(n+1, 100):
        if n % 10 != 0 and d % 10 != 0:
            n_str = [int(i) for i in str(n)]
            d_str = [int(i) for i in str(d)]

            for i in n_str:
                if i in d_str:
                    _n = n_str.copy()
                    _d = d_str.copy()
                    _n.remove(i)
                    _d.remove(i)
                    n_int = int(_n[0])
                    d_int = int(_d[0])

                    if (n/d == n_int/d_int):
                        prod *= n/d

print(Fraction('%.3f'%(prod)))