11.1. For some natural numbers $n>m$, the number $n$ turned out to be representable as the sum of 2021 addends, each of which is some integer non-negative power of the number $m$, and also as the sum of 2021 addends, each of which is some integer non-negative power of the number $m+1$. For what largest $m$ could this have happened (for some $n>m$)?

(A. Kuznetsov)
Answer. $m=2021$.

Solution. Let $m>2021$. Since any power of the number $m+1$ gives a remainder of 1 when divided by $m$, the sum of 2021 such powers gives a remainder of 2021 when divided by $m$. On the other hand, powers of the number $m$ give only remainders of 0 or 1 when divided by $m$, so the sum of 2021 powers of the number $m$ can give a remainder of 2021 when divided by $m$ only if all the terms are equal to 1. But then $n=2021m$, which in bases $m$ and $m+1$ when $m=2021$ has a digit sum of 2021 (thus, it also satisfies the condition of the problem): $n=m^{2}+m(m-1)=(m+1)^{2}+(m+$ $+1)(m-4)+3$