Let $A, B$ and $C$ be non-zero digits, so that $B C$ is a two-digit positive integer and $A B C$ is a three-digit positive integer made up of the digits $A, B$ and $C$. Suppose that

$$
\begin{array}{r}
B C \\
A B C \\
+\quad A B C \\
\hline 876
\end{array}
$$

What is the value of $A+B+C$ ?
We start by considering the ones (units) column of the given sum.

From the units column, we see that the units digits of $3 C$ must equal 6 .

The only digit for which this is possible is $C=2$. (We can check that no other digit works.)

Thus, the sum becomes

$$
\begin{array}{r}
B 2 \\
A B 2 \\
+\quad A B 2 \\
\hline 876
\end{array}
$$

We note that there is no "carry" from the ones column to the tens column.

Next, we consider the tens column.

From the tens column, we see that the units digit of $3 B$ must equal 7.

The only digit for which this is possible is $B=9$.

Thus, the sum becomes

$$
\begin{array}{r}
92 \\
A 92 \\
+\quad A 92 \\
\hline 876
\end{array}
$$

The "carry" from the tens column to the hundreds column is 2 .

Next, we consider the hundreds column.

We see that the units digit of $2 A+2$ (the two digits plus the carry) must equal 8 .

Thus, the units digit of $2 A$ must equal 6 . This means that $A=3$ or $A=8$.

The value of $A=8$ is too large as this makes $A 92=892$ which is larger than the sum.

Therefore, $A=3$.

(Putting this analysis another way, we have $92+(100 A+92)+(100 A+92)=876$. Simplifying, we obtain $200 A+276=876$ or $200 A=600$ and so $A=3$.)

We can check that $92+392+392=876$.) Therefore, $A+B+C=3+9+2=14$.

ANSWER: 14