Suppose $n\geq 3$ is an integer. There are $n$ grids on a circle. We put a stone in each grid. Find all positive integer $n$, such that we can perform the following operation $n-2$ times, and then there exists a grid with $n-1$ stones in it: $\bullet$ Pick a grid $A$ with at least one stone in it. And pick a positive integer $k\leq n-1$. Take all stones in the $k$-th grid after $A$ in anticlockwise direction. And put then in the $k$-th grid after $A$ in clockwise direction.
Problem
Source: 2022 Japan Junior MO Final P2
Tags: combinatorics, AZE CMO TST, AZE EGMO TST
Snark_Graphique
13.02.2022 01:22
All even $n$
Label the grids as $1, 2, 3, \dotsc, n$ clockwise. Note that whenever a stone from grid $i$ is moved to grid $j$, the number $i - j$ is even. Therefore, if $n$ is even, then the stone from grid with label $1$ can only go into grids with odd label and hence it would not be possible to collect $n-1$ stones into one grid. Next, if $n$ is odd, we can collect the stones in the following order: [asy][asy] import olympiad; size(7cm); pair A = dir(90); real[] L; for (int i=1; i<12; ++i) { dot(" ", dir(90-(360*i)/11), dir(90-(360*i)/11)); } for (int i=1; i<6; ++i) { draw(dir(90-(360*i)/11)--dir(90+(360*i)/11), MidArrow); } for (int i=2; i<6; ++i) { draw(dir(90+(360*i)/11)--dir(90-(360*(i-1))/11), MidArrow); } [/asy][/asy]
abdirasulov
31.01.2023 11:40
BUMP. It's a very interesting question, but I can't understand the answer. I think only prime numbers can be an answer, but not all of them.
Assassino9931
03.06.2024 02:55
Alternative argument for even $n$. If we colour the grids black and white in an alternating fashion, then if a stone is initially in black (white) grid, then it always stays in a black (white) grid. Therefore each grid can at any point contain no more than $\frac{n}{2} < n-1$ stones.