1. Determine all three-digit numbers $\overline{a b c}$ such that $\overline{a b c} \div c = \overline{b c}$ and $a \neq 0, b \neq 0$ and $c \neq 0$.
Solution: From $\overline{a b c}: c=\overline{b c}$ we get $\overline{a b c}=c \cdot \overline{b c}$, which means the product $c \cdot \overline{b c}$ is a number whose unit digit is $c$. This is only possible if $c \in\{1,5,6\}$.

$1^{\circ} c \neq 1$ because the product $1 \cdot \overline{b l}$ is a two-digit number.

$2^{\circ}$ For $c=5$ we have: $100 a+10 b+5=5 \cdot(10 b+5)$, which simplifies to $5 a=2 b+1$. This is only possible for $a=1, b=2$ and for $a=3, b=7$. The required numbers are 125 and 375.

$3^{\circ}$ For $c=6$ we have $100 a+10 b+6=6 \cdot(10 b+6)$, which simplifies to $10 a=5 b+3$. In this equation, the left side is divisible by 5, but the right side is not. There is no solution. Therefore, the required numbers are 125 and 375.