# Calculating p-adic valuation $v_p(n)$, using basic properties

Calculating p-adic valuation $v_p(n)$ I'm not confident with the properties of $v_p(n)$ Where $v_p(n) =$ the biggest integer $e$ such that $p^e$ divides $n$, if $n\not=0$, and $+\infty$ if $n=0$.

Would I be correct in calculating $v_3(50!/{25!25!})$ By calculating $v_3(50!)-v_3(25!)-v_3(25!)$ ?

$v_p(n) =$ the biggest integer $e$ such that $p^e$ divides $n$, if $n\not=0$, and $+\infty$ if $n=0$. For the rest you are right, as the function $v_p$ as defined "by me" is multiplicative : $v_p(nn') = v_p(n) v_p(n')$ for $n,n'$ non zero integers.
• thanks, as a follow up how could you evaluate say $v_{13}((13^5)!)$ Mar 3 '15 at 22:33
• $v_13((13^5)!)=\sum_{k=1}^{13^5} v_p (k)$. But you have a nice formula for $v_p(n!)$, look my answer here : math.stackexchange.com/questions/1127894/… Mar 3 '15 at 22:34
• You will note that writing $13^5$ in base $13$ is pretty easy. ;-) Mar 3 '15 at 22:40
• Thanks I appreciate the help, this is probably a silly question but why is $13^5$ in base 13 easy? Mar 3 '15 at 22:42
• It's silly if you're used to this, but if it's your first time, nothing is silly ! ;-) Writing $x\in\mathbf{N}^*$ in base $13$ means writing it $x = \sum_{k=0}^d a_i 13^k$, where the $a_i$'s are $\geq 0$ and $<13$. Then, $13^5 = 1\times 13^5$ is the expansion in base $13$ of $13^5$. Is it clear now ? Mar 3 '15 at 22:49