How to find the highest power of prime $p$ in $N!$, when



I know that the highest power of prime contained in $N!$ is given by:

$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$ But how to apply the restrictions?


The formula is true for all $N \geq 0$; the restriction does not affect its truth value. However, we can simplify things.

  • We see that $s_p(N!)=s_p((N-(N \text{ mod } p))!)$. Since the factors $N-(N \text{ mod } p)+1,N-(N \text{ mod } p)+2,\ldots,N$ make no contribution to $s_p(N!)$, as they are coprime to $p$. [Here $N \text{ mod } p$ represents the residue in $\{0,1,\ldots,n-1\}$.]

    E.g. in the first case, this means $s_p(N!)=s_p(p^r!)$.

  • We can "chop off" terms $\lfloor N/p^k \rfloor$ for which $N/p^k<1$, since the floor function will make them zero.

    So, when applying the formula to $p^r!$, we need only account for the terms $\lfloor p^r/p^k \rfloor$ where $1 \leq k \leq r$. Further, in this case, $p^k$ divides $p^r$, so the floor functions are unnecessary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.