Subject (3). Consider the set $A=\{\overline{a b} \mid \overline{a b}$ divisible by 5\}. For a subset $B \subseteq A$, we denote by $m_{B}$ the arithmetic mean of its elements. Determine the subsets $B$ of maximum cardinality, knowing that $m_{B} \in A$.

## TraiAN PredA
Solution. Considering the set $A^{\prime}=\{2,3, \ldots, 19\} \bigsqcup^{4}$, the problem reduces to finding subsets $B^{\prime} \subseteq A^{\prime}$ of maximum cardinality, such that the arithmetic mean of their elements is a natural number. Since $B^{\prime}=A^{\prime}$ does not work, as $\frac{2+3+\cdots+19}{18}=\frac{189}{18}$, we will look for sets $B^{\prime}$ with only 17 elements, so $B^{\prime}=A^{\prime} \backslash\{k\}$, for some $2 \leq k \leq 19$. Then $m_{B^{\prime}}=\frac{189-k}{17}$ must be a natural number, so $17 \mid k-2$, which is possible if and only if $k \in\{2,19\}$. It follows that $B=A \backslash\{10\}$ and $B=A \backslash\{95\}$.