TITLE: Dividing $2012!$ by $2013^n$
QUESTION [17 upvotes]: What's the largest power $n$ such that $2012!$ is divisible by $2013^n$?
It doesn't look like its divisible at all since $2012<2013$; am I right?

REPLY [12 votes]: We have $2013=3\times11\times61$. $2012!$ has $\left\lfloor\frac{2012}{3}\right\rfloor=670$ terms in the product that are multiples of $3$, $\left\lfloor\frac{2012}{11}\right\rfloor=182$ terms that are a multiple of $11$ and $\left\lfloor\frac{2012}{61}\right\rfloor=32$ terms in the product that are multiples of $61$. Therefore  $2012!$ only has $32$ "$61$"s that can be cancelled off by $61$ of $2013$. So the maximum value for $n$ is $32$.

REPLY [4 votes]: Maximum power of $61=$ in  $2012!=\left\lfloor\frac{2012}{61}\right\rfloor$ is $32$ ,maximum power of $11$ in $2012!$ is $61\times 3-1=\left\lfloor\frac{2012}{11}\right\rfloor$, maximum power of $11$ in $2012!$ is $11\times 61-1=\left\lfloor\frac{2012}{3}\right\rfloor$
So the minimum of these numbers will be the maximum power of $2013$ that divides $2013!$ 
I have used the fact that maximum power of $p$ a prime dividing $q!$ is $=\left\lfloor\frac{q}{p}\right\rfloor$.$\left\lfloor\frac{}{}\right\rfloor$ is the greatest integer function.