Let $k\ge 14$ be an integer, and let $p_k$ be the largest prime number which is strictly less than $k$. You may assume that $p_k\ge \tfrac{3k}{4}$. Let $n$ be a composite integer. Prove that if $n=2p_k$, then $n$ does not divide $(n-k)!$, if $n>2p_k$, then $n$ divides $(n-k)!$.
Problem
Source:
Tags: Divisibility Theory
14.09.2007 22:17
The solution on Kalva seems incomplete to me... at least I see no justification why we'd only have those 3 cases, e.g. why we cannot have $ q<r$. Can anyone explain me why it is obvious?
14.09.2007 22:44
Peter wrote: You may assume that $ p_{k}\ge\tfrac{3k}{4}$. And my question is if the statement above is always true I mean I know how to prove that $ p_{k}\geq\frac{k}{2}+3$ but that's about it ...
14.09.2007 23:25
There are many variants on that theorem. For sufficiently high $ a$ we have a prime above $ \frac{b-1}{b}a$, I believe.
19.01.2009 13:52
The link solutions on kalva appears not to work . Anyway I balieve that the assumtion quoted by valentin vornicu could be a wrong assution if we take into account large enough prime gaps or lagoons startin at some big enough n! +2.
19.01.2009 13:58
Phill: Use the Wayback Machine to access Kalva (if you live in GB, use a proxy like TOR to access the Wayback Machine...). And the bounds for $ p_k$ are correct, they are a rather known generalization of the Bertrand postulate. darij
19.01.2009 14:30
Thank you Darij