Question

The number of zeros at the end of $60!$ is:

Answer 1

The number of trailing zeros in the decimal representation of $n!$, the factorial of a non-negative integer n, can be determined with this formula:

$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+......+\frac{n}{5^n}$

where, $k$ much be chosen such that $5^{(k+1)}>n$

Given,

$60!$

$=\frac{60}{5}+\frac{60}{5^2}$

$=12+2=14$ zeros