2014!

wolframalpha. yolo.

One "0" at the end means that the prime factorization contains a "2" and a "5".

Since the number of "2"s in $2014!$ is clearly greater than the number of "5"s in it, we only have to find the number of "5"s in the prime factorization of $2014!$

$\begin{array}{cl} &\mbox{The number of "5"s in the prime factorization of 2014!}\\ =&\left\lfloor\frac{2014}5\right\rfloor+\left\lfloor\frac{2014}{25}\right\rfloor+\left\lfloor\frac{2014}{125}\right\rfloor+\left\lfloor\frac{2014}{625}\right\rfloor+\left\lfloor\frac{2014}{3025}\right\rfloor\\ =&402+80+16+3+0\\ =&\boxed{501} \end{array}$

This can be computed per the Factorial Trailing Zero formula:

$Z = \Sigma_{k=1}^{\lfloor \log_{5}(n) \rfloor} \lfloor \frac{n}{5^k} \rfloor$

For $n = 2014$ , we have $\Sigma_{k=1}^{\lfloor \log_{5}(2014) \rfloor} \lfloor \frac{2014}{5^k} \rfloor = \boxed{501}.$