Solve if each letter is a different digit.

ABA
+ CBC
CBCB
-

If two 3-digit numbers add up to a 4-digit number, the leading digit of the answer must be a 1, so C=1.

Now take the tens digit, B+B=1, then
B must be 0 or 5 when we consider the carry of 1 from the units digit.
If A+C=A+1 from the units digit has a carry, then A=9.
Since A+C equal B in both the units and thousand's digits, we conclude that there is not carry in the tens digit (B+B=C), so B=0.
The sum is thus:

 909
+ 101
-------
 1010

1010

To solve the given equation where each letter represents a different digit, follow these steps:

1. Start by assigning a value to one of the letters. Let's assume A = 1. In this case, C cannot be 1 because each letter must be different.

2. Since A + C is greater than 10 (ABA + CBC), we know that A must be either 8 or 9. Let's assume A = 9.

3. Now, let's look at the digit in the units place of the sum (CBCB). When added, A + C equals B (9 + C = B). Since B must be less than 10, C can only be 0 or 1. But since each letter must be different, C cannot be 1. Therefore, C = 0.

4. Now let's look at the second column from the right (ABA + CBC = CBCB). The column contains B, B, and C. From the previous step, we know that C = 0. So, B + 0 = B, which means B must be 0. However, each letter must be different, so B cannot be 0.

5. Since we've encountered a contradiction, the initial assumption that A = 1 was incorrect. Let's reconsider A = 1.

6. With A = 1, A + C = B, which means B must be C + 1. Since A + C cannot be greater than 10, B can only be 2. Therefore, B = 2.

7. Now, A + C = 2 (1 + C = 2). Solving this gives C = 1.

8. Finally, we have A = 1, B = 2, and C = 1. Substituting these values back into the equation, we have:

121
+ 212
212