If digits $A$, $B$, and $C$ (between $0$ and $9$ inclusive) satisfy
    \begin{tabular}{c@{\,}c@{\,}c@{\,}c}
      & $C$  & $C$ & $A$ \\
    + & $B$ & $2$ & $B$ \\\hline
      & $A$ & $8$ & $8$ \\
    \end{tabular}
    what is $A \cdot B \cdot C$?


[i]2021 CCA Math Bonanza Individual Round #5[/i]
To solve the problem, we need to determine the digits \(A\), \(B\), and \(C\) that satisfy the given addition problem:

\[
\begin{array}{c@{\,}c@{\,}c@{\,}c}
  &  C  &  C  &  A  \\
+ &  B  &  2  &  B  \\ \hline
  &  A  &  8  &  8  \\
\end{array}
\]

We will analyze the problem column by column, starting from the rightmost column.

1. **Rightmost Column:**
   \[
   A + B = 8 \quad \text{or} \quad A + B = 18
   \]
   Since \(A\) and \(B\) are digits (0-9), the sum \(A + B = 18\) is not possible because it would require a carry, which is not indicated in the problem. Therefore, we have:
   \[
   A + B = 8
   \]

2. **Middle Column:**
   \[
   C + 2 = 8 \quad \text{or} \quad C + 2 = 18
   \]
   Again, since \(C\) is a digit, \(C + 2 = 18\) is not possible. Therefore, we have:
   \[
   C + 2 = 8 \implies C = 6
   \]

3. **Leftmost Column:**
   \[
   C + B = A \quad \text{or} \quad C + B = A + 10
   \]
   Since \(C = 6\), we substitute:
   \[
   6 + B = A \quad \text{or} \quad 6 + B = A + 10
   \]
   Given \(A + B = 8\), we substitute \(A = 8 - B\) into the equation:
   \[
   6 + B = 8 - B \quad \text{or} \quad 6 + B = 8 - B + 10
   \]
   Solving the first equation:
   \[
   6 + B = 8 - B \implies 2B = 2 \implies B = 1
   \]
   Substituting \(B = 1\) into \(A = 8 - B\):
   \[
   A = 8 - 1 = 7
   \]

4. **Verification:**
   We now have \(A = 7\), \(B = 1\), and \(C = 6\). We verify the original addition problem:
   \[
   \begin{array}{c@{\,}c@{\,}c@{\,}c}
     &  6  &  6  &  7  \\
   + &  1  &  2  &  1  \\ \hline
     &  7  &  8  &  8  \\
   \end{array}
   \]
   The addition is correct, confirming our solution.

The final answer is \(\boxed{42}\).