12th IMO

mousavi wrote: 12th IMO http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366692&sid=518ba005ce9ea6e629a496ffa278b3a6#p366692 It is different

$(n+4)|n(n+1)(n+2)(n+3)(n+5)(n+6)(n+7)(n+8)$, so $(n+4)|576$. At least one of $n,n+1,\ldots,n+8$ is a multiple of 7, so at least two of them are multiples of 7, hence we get $n\equiv0,1\pmod7$. Therefore $n+4$ is a divisor of 576 of the form $7k+4$ or $7k+5$, we get $n+4\in\{12,18,32,96,144\}$. But we must have $(n+3)|n(n+1)(n+2)(n+4)(n+5)(n+6)(n+7)(n+8)$, so $n+3|720$. However none of $11,17,31,95,143$ divides $720$. Thus no such $n$ exists.

No, if $a|b$ and $c|b$ it does not follow that $a|b$, which is what you use in your first line. One can observe that the one group will have 5 factors and the other one only 4, thus asymptotically it cannot hold, which gives a(very low) bound on $n$. More precisely, if we have $n>\frac{(n+5)(n+6)(n+7)(n+8)}{(n+1)(n+2)(n+3)(n+4) }$(if n>4, then this holds whenever n>16), then $ \textup{group with 5 factors} \geq n(n+1)(n+2)(n+3)(n+4) > (n+5)(n+6)(n+7)(n+8) \geq \textup{group with 4 factors},$ a contradiction. So $n$ is smaller or equal then 16. Now it's down to check these options. It is of help to see that each group should have a factor 7, which implies $n=-1,0 \mod 7$, and also with 5, which gives $n\neq 1 \mod 5$.

Korneelmans wrote: No, if $a|b$ and $c|b$ it does not follow that $a|b$, which is what you use in your first line. No I didn't. I used modulo (n+4).

Korneelmans,what do you mean by $a|b,c|b$ means $a|b$? Will it be $a|c$?