The 3-digit number $A A A$, where $A \neq 0$, and the 6 -digit number $A A A B B B$ satisfy the following equality: $A A A \times A A A+A A A=A A A B B B$.
G8.3 Find $A$.
G8.4 Find $B$.
$$
\begin{array}{l}
A(111) \times A(111)+A(111)=A(111000)+B(111) \\
111 A^{2}+A=1000 A+B
\end{array}
$$

Consider the thousands digit: $9<A^{2} \leq 81$
$$
\Rightarrow A=4,5,6,7,8,9
$$

When $A=4: 111 \times 16+4=4000+B$ (rejected)
When $A=5: 111 \times 25+5=5000+B$ (rejected)
When $A=6: 111 \times 36+6=6000+B$ (rejected)
When $A=7: 111 \times 49+7=7000+B$ (rejected)
When $A=8: 111 \times 64+8=8000+B$ (rejected)
When $A=9: 111 \times 81+9=9000+B$
$$
\begin{array}{l}
\therefore A=9 \\
B=0
\end{array}
$$