Monday, September 17th

Problem

In how many zeros does $2012!$ end?

Solution

Since each zero comes from multiplying a 2 with a 5, it suffices to count how many 5 factors are in $2012!$.  Hence the number of zeros is

$\sum_{k=1}^\infty \left\lfloor\frac{2012}{5^k}\right\rfloor=501$