$28 \cdot 2$ In the following expression, different letters represent different digits, then $C$ is
\begin{tabular}{rrr}
$A$ & $B$ & $C$ \\
$A$ & $B$ \\
+ & & $A$ \\
\hline \multirow{3}{3}{$\quad 0$} & 0
\end{tabular}
(A) 1 .
(B) 3 .
(C) 5 .
(D) 7 .
(E) 9 .
(7th American Junior High School Mathematics Examination, 1991)
[Solution 1] From the equation, we can see that $A=3$ or 2 (if there is a carry, it can be at most 1).
$\because B+A$ plus the carry digit from $A+B+C$ (which can be at most 2) and the last digit is 0, it will carry over, so $A=2$.

Given $A=2$, we know $A+B+C \leqslant 20$, and the carry does not exceed 2. Therefore, from the last digit of $B+A$ being 0, we have $8 \leqslant A+B \leqslant 10$.
$\because B+A+C$ has a last digit of 0, and $8 \leqslant A+B \leqslant 10$,
$$
\therefore \quad A=2, \quad B=7, \quad C=1 \text {. }
$$

Therefore, the answer is $(A)$.
[Solution 2] The given equation can be written as $111 A+11 B+C=300$,
$\because 0<11 B+C<11 \cdot 9+9=108$,
$\therefore A=2$, so $11 B+C=78$.
Given $0 \leqslant C \leqslant 9$, then $69 \leqslant 11 B \leqslant 78$.
Hence $B=7$, and thus $C=1$.
Therefore, the answer is $(A)$.