Problem 795

The number of zeroes that $n!$ ends with is the largest power of 10 dividing $n!$. The exponent of 5 dividing $n!$ exceeds the exponent of 2 dividing $n!$, so we simply seek the exponent of 5 dividing $n!$. For a number less than 125, this exponent is just the number of multiples of 5, but not 25, less than $n$ plus twice the number of multiples of 25 less than $n$. Counting up, we see that 24! ends with 4 zeroes while 25! ends with 6 zeroes, so $n!$ cannot end with 5 zeroes. Continuing to count up, we see that the smallest $n$ such that $n!$ ends with 10 zeroes is 45.