Problem 1741

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 $k \equiv 3^{-1} \cdot (b-1) \pmod{n}$ implies $3k+1 \equiv b \pmod{n}$ 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. $\square$