Dividing $2012!$ by $2013^n$

Question

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?

http://www.proofwiki.org/wiki/Factorial_Divisible_by_Prime_Power — lab bhattacharjee, Jun 13 '13 at 17:11
http://math.stackexchange.com/questions/20737/prime-powers-that-divide-a-factorial — lab bhattacharjee, Jun 13 '13 at 17:11
Also http://math.stackexchange.com/questions/363738/prove-the-converse-of-wilsons-theorem — lab bhattacharjee, Jun 13 '13 at 17:13
You would be right if $2013$ was prime. But $2013=3\times 11\times 61$. — Start wearing purple, Jun 13 '13 at 17:13
I forgot you can factor 2013, how do you know if 2013 is prime or not? — Jong Sui, Jun 13 '13 at 17:14
You can tell right away that its divisible by 3 if you add the digits in the number (2+0+1+3=6), and the result is divisible by 3. Since 6/3=2, which is a integer, then you know 2013 is divisible by 3 as well. — Sujaan Kunalan, Jun 13 '13 at 17:15
That only works for $3$ and $9$ for summing up the digits to check for divisibilities — , Jun 13 '13 at 17:18
@Bitwise How is 9! divisible by 100 (10^2)? 9! is divisible by 10^n only for n=0..1 — SinisterMJ, Jun 14 '13 at 08:16

score 46 · Accepted Answer · edited Jun 13 '13 at 20:29

46

$2013=3 \times 11 \times 61$. 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.

edited Jun 13 '13 at 20:29

jscs

107

answered Jun 13 '13 at 17:20

Alexander

4,292

score 12 · Answer 2 · answered Jun 13 '13 at 17:25

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 $32$.

score 4 · Answer 3 · answered Jun 13 '13 at 17:25

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.

score 2 · Answer 4 · answered Jun 13 '13 at 17:20

2

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 $n=32$.

answered Jun 13 '13 at 17:20

Start wearing purple

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Dividing $2012!$ by $2013^n$

4 Answers4