BMO 2022 shortlist

Solution 1. Write $a=dr$ and $b=ds$ where $d=\gcd (a,b)$ and $(r,s)=1$. Then $d^{2021}{r}^{2021}{s}^{2021}$ divides $n$. Furthermore $2\cdot 3\cdot 337\mid d(r-s)$.

Case 1: Assume $337\mid d$. Since $(r,s)=1$, we may assume that $r$ is odd. Then $$\{337r^2,337r^4,\dots ,337r^{10},337r^{12}\}$$ works. Indeed each of these six divisors of $n$ is congruent to $1\mathrm{mod}4$ so their sum is a multiple of $2$ but not of $4$. If $3\mid r$ then the sum is $0\mathrm{mod}3$ while if $3\nmid r$ then each of these divisors is $1\mathrm{mod}3$ so the sum is again $0\mathrm{mod}3$. Therefore the sum is a multiple of $2022$ but not of $2022^2$.

Case 2: Assume $337\nmid d$. Then $337\mid r-s$ and since $(r,s)=1$ then $337\nmid rs$. Consider the $2022^2$ divisors of $n$ of the form $r^k{s}^{\ell }$ where $k,\ell \in \{0,1,2,\dots ,2021\}$. Since none of them is a multiple of $337$, they have at most $2\cdot 3\cdot 336$ distinct remainders modulo $2022$. Therefore at least $\frac{2022^2}{2\cdot 3\cdot 336}>2022$ of them have the same remainder modulo $2022$. Pick $2023$ out of those, say $d_1,d_2,\dots ,d_{2023}$. Let $S$ be their sum. We claim that there is a subset of $2022$ of them that will work. Note that the sum of any such subset is a multiple of $2022$. It is enough to show that there is such a subset whose sum is not divisible by $337^2$. If this is not the case then $S-d_i\equiv 0\mathrm{mod}337^2$ for each $i=1,2,\dots ,2023$. In particular, all $d_i$ are congruent mod $337^2$. Say that $d_i\equiv k\mathrm{mod}337^2$ for each $i$. Then $337\nmid k$ and so the sum of any $2022$ of them is congruent to $2022k\ne 0\mathrm{mod}337^2$.

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Alternative solution.

We start with the following claim: Claim. If $k$ is a positive integer, then $a^k{b}^{2021-k}\mid n$. Proof of the Claim. We have that $n^{2021}=n^k\cdot n^{2021-k}$ is divisible by $a^{2021k}\cdot b^{2021(2021-k)}$ and taking the $2021$-root we get the desired result. $\square$

Back to the problem, we will prove that the set $T=\{a^k{b}^{2021-k},k\ge 0\}$ consisting of $2022$ divisors of $n$, has the desired property. The sum of its elements is equal to $$S=\sum _{k=0}^{2021}{a}^k{b}^{2021-k}\equiv \sum _{k=0}^{2021}{a}^{2021}\equiv 0\mathrm{mod}2022.$$ On the other hand, the last sum is equal to $\frac{a^{2022}-b^{2022}}{a-b}$. We will prove that this is not divisible by $9$. Indeed, if $3^t\mid a-b$ then, since $3^1\mid 2022$, by the Lifting the Exponent Lemma, we have that $3^{t+1}\mid a^{2022}-b^{2022}$. This implies that $S$ is not divisible by $9$, therefore, $2022^2$ doesn't divide $S$.