Project Euler, Problem 33: Digit Cancelling Fractions

The fraction $\frac{49}{98}$ is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that $\frac{49}{98} = \frac{4}{8}$ , which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, $\frac{30}{50} = \frac{3}{5}$ , 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.

I realized that the common number in the numerator and the denominator had to occupy the 10s place in the denominator and the 1s place in the numerator in order to satisfy the requirement that the fraction be less than 1.

Then, I iterated through three numbers: The number common to both the numerator and the denominator (which I called the anchor), the 10s place in the numerator, and the 1s place in the denominator.

I ignored certain values and saved the fractions that fit the criteria, multiplied them, then reduced them.

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