How many trailing zeros are in 125! ?
Answer: C
The number of trailing zeros in a factorial n! is determined by the number of times 5 is a factor. We use the formula: \(\lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + ...\)
For 125!, this is \(\lfloor 125/5 \rfloor + \lfloor 125/25 \rfloor + \lfloor 125/125 \rfloor = 25 + 5 + 1 = 31\).
There are 31 trailing zeros in 125!.