Compute the product of all positive integers $n$ for which $3(n!+1)$ is divisible by $2n - 5$.
Problem
Source: Netherlands TST for IMO 2017 day 3 problem 3
Tags: number theory
Math-Ninja
01.02.2018 18:20
In an inequality format?
Medjl
01.02.2018 20:30
I am sure that i saw something similar in a turkey Mo .
programjames1
01.02.2018 22:55
Borbas wrote:
Show that if $n\ge4$ then $2n-5$ is prime.
Do you mean $2n-5$ must be prime for the condition to hold? Because not all integers $2n-5$ are prime. EDIT: Yeah, pretty sure you do mean that.
If $n=4,5$, then $2n-5$ is prime. If $n>5$ then $2n-5> n$. If $2n-5$ is not prime, then there exists some two $x,y$ such that $x,y|n!$ and $x,y|2n-5$ (as the smallest $d$ is $2$ and $2n-5<2n$). But then, $3(n!+1)\equiv 3\pmod{x,y}$ and only $1$ of $x,y$ can be $3$, so $2n-5\nmid 3(n!+1)$
Then, $2n-5$ must be prime.
onlygeo
23.02.2018 12:21
Medjl wrote: I am sure that i saw something similar in a turkey Mo . https://artofproblemsolving.com/community/c6h536626p3082205
MazeaLarius
23.02.2018 13:21
N=3k+1 (the problem should be ready right now)
Opsthisisatypo
22.11.2020 03:38
You should finish it by applying Wilson's theorem and a little casework