12.

Every zero at the end of a number corresponds to one of the factors being 10 (e.g., 23000 has 3 factors of 10 (i.e. $10^3$ divides 23000). But 10 itself can be factorised into $5 \times 2$, so we're looking for the smallest power of 2 or 5 and that will be our answer. $50!$ has at least 25 factors of 2, plenty more than we expect it to have factors of 5. Of the numbers 1 to 50, ten of them (5, 10, ..., 50) have 5 as a factor, and two of those have two factors of 5. So $50!$ has 12 factors of 5 (less than 25, good). Hence $50!$ has 12 zeros at its end.

*This problem is taken from the UKMT Mathematical Challenges.*

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