A student had $18$ papers. He seleced some of these papers, then he cut each of them in $18$ pieces.He took these pieces and selected some of them, which he again cut in $18$ pieces each.The student took this procedure untill he got tired .After a time he counted the pieces and got $2012$ pieces .Prove that the student was wrong during the counting.
Problem
Source: Kosovo IMO TST 2012 problem 1
Tags: modular arithmetic, combinatorics proposed, combinatorics
VIPMaster
23.04.2012 00:00
I'm not sure if I understand this question completely, but:
It's clear that the number of pieces increases by $17$ each time he cuts things. Thus the number he ends up with should be $17$ (mod $18$), which $2012$ is not (it leaves a remainder of $14$), so the student was wrong.
uglysolutions
27.04.2012 18:41
VIPMaster, you said it the other way round: the number of pieces is always $\equiv 18 \equiv 1 \pmod{17}$, whilst $2012 \equiv 6 \not \equiv 1 \pmod{17}$.
yugrey
27.04.2012 20:57
This does seem a little easy for an olympiad TST, even a #1... Also, isn't it Number Theory if it is modular arithmetic?
Arberi1
27.04.2012 21:07
Yes it's number theory , but I see that I posted it accidentaly here
tsiannis
19.03.2022 22:46
We observe that in each procedure there are added $17k$ papers. So, the result must be $1 (mod 17)$ since we started with $18$. $2012 \equiv -1 (mod 17)$ so it is not the case.