how many zeroes does 2012! have at the end? 
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? 


How many zeroes does $2012!$ end with? 

My idea is: 
402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625$ 
How can we "show" that this is true? 

The correct answer is 501.
In order to find the number of zeros is same as finding the number of factors of powers of $5$. There are more factors of powers of $2$ than the factors of powers of $5$.
For instance $10! = 3628800 = \hspace{3pt}2^8 \hspace{3pt} 3^4 \hspace{3pt}5^2\hspace{3pt} 7$
$$\left \lfloor \frac{n}{p}  \right \rfloor +\left \lfloor \frac{n}{p^2}   \right \rfloor +\left \lfloor \frac{n}{p^3}\right  \rfloor  + \cdots \left \lfloor \frac{n}{p^{k-1}}  \right \rfloor$$
where $\left \lfloor \frac{n}{p^k} \right \rfloor=0$. 
In this case $k=5$ because $\left \lfloor \frac{2012}{5^5} \right \rfloor=0$
$$\left \lfloor \frac{2012}{5}  \right \rfloor =402, \hspace{6pt} \left \lfloor \frac{2012}{5^2}  \right \rfloor = \left \lfloor \frac{402}{5}  \right \rfloor =80$$
$$\left \lfloor \frac{2012}{5^3} \right \rfloor = \left \lfloor \frac{80}{5}  \right \rfloor =16, \hspace{6pt} \left \lfloor \frac{2012}{5^4} \right \rfloor = \left \lfloor \frac{16}{5}  \right \rfloor =3$$
$$\left \lfloor \frac{2012}{5}  \right \rfloor+ \left \lfloor \frac{2012}{5^2}\right \rfloor +\left \lfloor \frac{2012}{5^3}\right \rfloor +\left \lfloor \frac{2012}{5^4} \right \rfloor = 402+80+16+3=501$$
