APMO 2004

[SANZEED]

Prove that
$\left\lfloor \frac{(n-1)!}{n(n+1)}\right\rfloor$
is even for every positive integer $n$.

[zadid xcalibured]

we can demonstrate two cases.one of n and n+1 is prime.then the other case is both of them are composite.im not done with the second case.

[*Mahi*]

It is quite easy(and quite cool ).
My solution scheme:

Step 1. : $\left\lfloor \frac{(n-1)!}{n(n+1)}\right\rfloor$=$\left\lfloor \frac{(n-1)!}{n} - \frac {(n-1)!}{n+1}\right\rfloor$
Step 2. : What Zadid said.
Step 3. : For none of $n,n+1$ prime , prove $n \text { and } n+1 | (n-1)!$ for "sufficiently large" $n$.
Step 3. : In that case,$\frac{(n-1)!}{n}$ and $\frac{(n-1)!}{n+1}$ has the same pairity.

[Phlembac Adib Hasan]

Very easy.I also solved this in national camp (while watching the exciting game of India vs Bangladesh. )

For composite $n$ it's very easy to prove.To prove for prime $p$, I just used this : $\frac {(p-1)!}{p+1}\equiv -1(mod\; p)$ and $\frac {(p-2)!}{p-1}\equiv -1(mod\; p)$.

[FahimFerdous]

A classic one! I solved it too during National Camp, with regards to Mahi who gave me a hint.

[Phlembac Adib Hasan]

Step 3. : For none of $n,n+1$ prime , prove $n \text { and } n+1 | (n-1)!$ for "sufficiently large" $n$.

"sufficiently large" শুনতেই বিশাল বিশাল সংখ্যার কথা মনে হচ্ছে।আসলে $n\ge 8$ নিলেই হয়।

[*Mahi*]

Step 3. : For none of $n,n+1$ prime , prove $n \text { and } n+1 | (n-1)!$ for "sufficiently large" $n$.

"sufficiently large" শুনতেই বিশাল বিশাল সংখ্যার কথা মনে হচ্ছে।আসলে $n\ge 8$ নিলেই হয়।

Yeah, and that's why I used the wink .