7.3. Determine all digits $(a, b, c, m, n)$ for which $\overline{0,(a b c)} \cdot \overline{3,(m n)}=3$.
## Solution:

Since $\overline{0,(a b c)}=\frac{\overline{a b c}}{999}$ and $\overline{3,(m n)}=\frac{\overline{3 m n}-3}{99}=\frac{297+\overline{m n}}{99}, \quad$ we obtain $\quad \overline{0,(a b c)} \cdot \overline{3,(m n)}=3 \Leftrightarrow \frac{\overline{a b c}}{999} \cdot \frac{297+\overline{m n}}{99}=3 \Leftrightarrow \overline{a b c} \cdot(297+\overline{m n})=3^{6} \cdot 11 \cdot 37 \Rightarrow 3^{6} \cdot 11 \cdot 37: \overline{a b c}$. Since $297+\overline{m n} \leq 297+$ $98=395 \Rightarrow \overline{a b c} \geq 752$. From $3^{6} \cdot 11 \cdot 37 \vdots \overline{a b c}$ and $\overline{a b c} \geq 752$ we get $\overline{a b c}=891$. Thus, $297+\overline{m n}=$ $333 \Rightarrow \overline{m n}=36$.

Answer: $a=8, b=9, c=1, m=3, n=6$.