BxMO Team Selection Test, March 2020

Suppose $n$ is not a multiple of 3 and that we have a stable assignment of the numbers $a_1,a_2,\dots ,a_n$, in that order on the circle. Then we have $a_i{a}_{i+1}{a}_{i+2}=n$ for all $i$, where the indices are considered modulo $n$. Hence, $$a_{i+1}{a}_{i+2}{a}_{i+3}=n=a_i{a}_{i+1}{a}_{i+2},$$ which yields $a_{i+3}=a_i$ (as all numbers are positive). Through induction, we find that $a_{3k+1}=a_1$ for all integers $k\ge 0$. Because $n$ is not a multiple of 3, the numbers $3k+1$ for $k\ge 0$ take on all values modulo $n$: indeed, 3 has a multiplicative inverse modulo $n$, hence $k\equiv 3^{-1}\cdot (b-1)(\mathrm{mod}n)$ implies $3k+1\equiv b(\mathrm{mod}n)$ for all $b$. We conclude that all numbers on the circle must equal $a_1$. Hence, we have $a_1^3=n$, where $a_1$ is a positive integer. Hence, if $n$ is not a multiple of 3, then $n$ must be a cube.

If $n$ is a multiple of 3, then we put the numbers $1,1,n,1,1,n,\dots$ in that order on the circle. In that case, the product of three adjacent numbers always equals $1\cdot 1\cdot n=n$. If $n$ is a cube, say $n=m^3$, then we put the numbers $m,m,m,\dots$ on the circle. In that case, the product of three adjacent numbers always equals $m^3=n$.

We conclude that a stable assignment exists if and only if $n$ is a multiple of 3, or a cube. Now we have to count the number of such $n$. The multiples of 3 with $3\le n\le 2020$ are $3,6,9,\dots ,2019$; these are $\frac{2019}{3}=673$ numbers. The cubes with $3\le n\le 2020$ are $2^3,3^3,\dots ,12^3$, because $12^3=1728\le 2020$ and $13^3=2197>2020$. These are 11 cubes, of which 4 are divisible by 3, hence there are 7 cubes which are not a multiple of 3. Altogether, there are $673+7=680$ values of $n$ satisfying the conditions. $\square$