Solution. From $a \cdot c=c$ it follows that $c=0$ or $a=1$. If $c=0$, then $a \cdot b \cdot b \cdot c \cdot d=0$, which is not possible. Therefore, $a=1$. From $b \cdot c=\overline{1 c}$ we get $(b-1) c=10$. From this, we get that $c \in\{1,2,5\}$. It is clear that $c \neq 1$ (since $a=1$). If $c=2$, then $b=6$. But then $a \cdot b \cdot b \cdot c \cdot d=612$, which is not possible since 17|612. If $c=5$, then $b=3$. It follows that $3 \cdot 3 \cdot 5 \cdot d=315$, i.e., $d=7$. Therefore, the required numbers are $13357,315,15$, 5.
Problem 5. Given four numbers: $\overline{a b b c d}, \overline{b a c}, \overline{a c}, c$ (where $a, b, c$ are different digits). Starting from the second, each number is equal to the product of the digits of the previous one. Determine the numbers $\overline{a b b c d}, \overline{b a c}, \overline{a c}, c$.