Dividing $2012!$ by $2013^n$

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?

Solution 1:

$2013=3 \times 11 \times 61$. Thirty-two naturals $ \leq 2012$ are divisible by $61$ and none are divisible by $61^2$. At least thirty-two naturals are divisible by $11$ and by $3$, and so we have that $2013^{32}$ divides $2012!$ but no larger power of $2013$ does.

Solution 2:

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 $32$.

Solution 3:

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.

Solution 4:

A pedestrian approach would be to ask WolphramAlpha to determine prime factors of $2012!$. And since $2013=3\times11\times61$, the answer to your question is $n=32$.