Bariz - Problem

Source: Nigerian Mathematics olympiad 2021 round 2 problem 2

Tags: combinatorics

Ejaifeobuks

15.02.2021 02:31

$N$ boxes are arranged in a circle and are numbered $1,2,3,.....N$ In a clockwise direction. A ball is assigned a number from${1,2,3,....N}$ and is placed in one of the boxes.A round consist of the following; if the current number on the ball is $n$, the ball is moved $n$ boxes in the clockwise direction and the number on the ball is changed to $n+1$ if $n<N$ and to $1$ if $n=N$. Is it possible to choose $N$, the initial number on the ball, and the first position of the ball in such a way that the ball gets back to the same box with the same number on it for the first time after exactly $2020$ rounds

I'm pretty sure my answer is wrong because it feels too easy... but anyway The answer is no, it is not possible. Suppose it was possible, observe that considering the box numbers mod $N$, we see that for the condition to hold and let $n$ be the ball number, $n + n+1 + n+2 + .... + n+k \equiv 0 \pmod N$ must imply that $k \equiv -1 \pmod {2020}$ since it must be that this procedure is periodic every $2020$ moves. But this means that $(k+1)(2n+k) \equiv 0 \pmod N$ and so $k \equiv -1 \pmod N$. But since $k \equiv -1 \pmod {2020}$, this means $N = 2020$. But then, checking, we see that $2020n + \frac{(2020)(2021)}{2} \equiv 0 \pmod {2020}$ means that $(1010)(2021) \equiv 0 \pmod {2020}$, which is false. Therefore, it is not possible to choose $N, n$ and the position such that the condition holds.