Task 2. Determine all possible values of the digits $a$ and $b$, such that the product of the numbers $\overline{54 a}$ and $\overline{63 b 1}$ is divisible by 12.
Solution. For the product $\overline{54 a} \cdot \overline{63 b 1}$ to be divisible by 12, it must be divisible by both 3 and 4. The number $\overline{63 b 1}$ is an odd number for any choice of the digit $b$, so it cannot be divisible by 4, which means the number $\overline{54 a}$ must be divisible by 4. According to the divisibility rule for 4, the number $\overline{54 a}$ is divisible by 4 when the two-digit ending $\overline{4 a}$ is divisible by 4, which is only true when $a=0, a=4$, or $a=8$.

If $a=0$, then $\overline{54 a}=540$ is a number divisible by 3, so the digit $b$ can be any digit.

If $a=4$ or $a=8$, then the number $\overline{54 a}=544$, or the number $\overline{54 a}=548$ is not divisible by 3, so the product $\overline{54 a} \cdot \overline{63 b 1}$ is divisible by 12 if the number $\overline{63 b 1}$ is divisible by 3. This means the sum of its digits $6+3+b+1=10+b$ must be divisible by 3, which is satisfied if $b=2, b=5$, or $b=8$.

According to this, the product $\overline{54 a} \cdot \overline{63 b 1}$ is divisible by 12 if $a=0$ and $b \in\{0,1,2,3,4,5,6,7,8,9\}$ or if $a \in\{4,8\}$ and $b \in\{2,5,8\}$.