0
$\begingroup$

I'm doing some exercises and I've read that, if $\alpha$ is the first prime factor of a number $m \geq 2$, then, for every $k \geq \alpha$, it is true that $gcd(k!\ mod\ m,\ m) > 1$.

I can see that $gcd(k!,\ m) > 1$ is true because $k!$ contains $\alpha$ as one of its factors, but I can't see where that mod m fits in.

$\endgroup$
  • $\begingroup$ Is $n$ and $m$ the same? $\endgroup$ – Hagen von Eitzen Apr 9 '14 at 21:35
  • $\begingroup$ Yes, I mistyped it, editing now $\endgroup$ – Ivan Apr 9 '14 at 21:36
2
$\begingroup$

By the division algorithm, $k!=qm+r$ for $q,r\in\mathbb{Z}$ with $0\le r<m$. Since $\alpha|k!$ and $\alpha|qm$, we must have that $\alpha|r$, and hence $\textrm{gcd}(k!\mod m,m)=\textrm{gcd}(r,m)>1$.

$\endgroup$
0
$\begingroup$

By Euclid $\ \gcd(k!\ {\rm mod}\ m,\, m) = \gcd(k!,m),\,$ and $\ \alpha\mid k!,\,m\,\Rightarrow\,\alpha\mid\gcd(k!,m)$

$\endgroup$

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.