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The number of trailing zeros at the end of $50!$ is
A12
B13
C11
D10
Answer & Solution
Correct answer: A. 12
Trailing zeros are governed by the power of 5 in $50!$: $\lfloor 50/5 \rfloor + \lfloor 50/25 \rfloor = 10 + 2 = 12$. (Powers of 2 are more plentiful, so 5 is the limiting factor.)
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