Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\mathbb{P}$ is the prime numbers set.

$p \in \mathbb{P}$

$a,b,c \in \mathbb{N}$

$n=a p^b+c$ where

$c= n\bmod p$

$b$ is the highest power of $p$ who divides $n-c$

How to find $\beta$ where $\beta$ is the highest power of $p$ who divides $n!$?

And how to find $\alpha$ where $\alpha$ is $\dfrac{n!}{p^\beta}\bmod p$?

$n!=\alpha p^\beta$

Probably the answer will appear $a$, $b$, $c$, $p$ and factorials.

Obs.:

$x = y\bmod z \iff x \equiv y \pmod{z}$ and $0 \leq x \leq z-1$

$x,y,z \in \mathbb{N}$

share|improve this question
    
What does "$c=\bmod (n,p)$" mean? –  Arturo Magidin Nov 8 '11 at 5:46
    
$c \equiv n \bmod p$ –  GarouDan Nov 8 '11 at 5:47
2  
I'm not familiar with that notation; I do not think it is standard. If you mean that $c$ is the remainder of dividing $n$ by $p$, sometimes this is written $n\bmod p$ (n\bmod p), but I think you'll have to spell it out in the body to make it clear. In any case, use \bmod instead of \mod, to get the right phrasing. –  Arturo Magidin Nov 8 '11 at 5:51
2  
I expect you are familiar with the fact that $\beta=\lfloor \frac{n}{p}\rfloor + \lfloor \frac{n}{p^2}\rfloor +\lfloor \frac{n}{p^3}\rfloor+\cdots$. We get a nice simple formula if $1\le a<p$ and $0 \le c<p$, and something a bit messier otherwise. –  André Nicolas Nov 8 '11 at 5:53
    
I'm using the Mathematica notation Mod[n,p] is the remainder of dividing $n$ by $p$... But if you think will be more clear change it, please do it for me =) –  GarouDan Nov 8 '11 at 5:55
show 6 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.