2015!

To help understand the reasoning behind it, consider: how do trailing zeros exist? When a $2$ and $5$ are multiplied, of course. There's always more $2$ 's than $5$ 's, though, so there are as many trailing zeros as there are $5$ 's. Powers of $5$ contain multiple $5$ 's, so you need to count them for as many $5$ 's as their power: $125,$ for example, counts as three $5$ 's. Now here's the general expression for number of trailing zeros: $Z(x) = \sum_{n=1}^{\lfloor\log_5(x)\rfloor}\lfloor\frac{x}{5^n}\rfloor$ Substitute $x = 2015$ to get $Z(2015) = \sum_{n=1}^{4}\lfloor\frac{2015}{5^n}\rfloor = \boxed{502}$