Q:

How much zeros has the number $1000!$ at the end?

I know that it depends of the factors of five and two.
But the number is too long to figure how much factos of five and two there are.
Any hints?

A:

There are always twos a-plenty. The exponent of prime $p$ occuring in $n!$ is well-known to be
$$ \lfloor n/p\rfloor +\lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor +\ldots$$
hence for $n=1000$ and $p=5$ we find
$$ \lfloor 1000/5\rfloor +\lfloor 1000/25\rfloor + \lfloor 1000/125\rfloor +\lfloor 1000/625\rfloor + \ldots= 200+40+8+1+0+\ldots=249$$
(Just to check, for $p=2$ we get
$$500+250+125+62+31+15+7+3+1\gg249 $$
so really more than enough)

