### Goodbye, 2021.

It is the last moment of 2021. The very last. Do your last farewells. I thank God and my parents for all the opportunities they have given me during this year. And I hope me and everyone get even better ones in the coming year. After this, no more 2021. I would like to share a meme at this moment. On this occasion, I am solving 2021 IMO #6; as the last of 2021. Let $m\ge 2$ be an integer, $A$ a finite set of integers (not necessarily positive) and $B_1,B_2,...,B_m$ subsets of $A$. Suppose that, for every $k=1,2,...,m$, the sum of the elements of $B_k$ is $m^k$. Prove that $A$ contains at least $\dfrac{m}{2}$ elements. Let $A=\{a_1, a_2, a_3, \dots, a_n\}$. Note that for some number $0 \leq N \leq m^{m+1} - m$ with $m | N$, we can choose integers $x_{1}, x_{2}, \ldots x_{m}$ so that $$0 \leq x_{i} < m$$ and \[N = x_{1}m + x_{2}m^{2} + \ldots + x_{m}m^{m}.\] We know this by dividing both sides by $m$ and then writing $N$ in base $m$. Next, notice that we can write $N$ as the sum