# Solution

What we have is: C + B = A. Then, we have to add B and C; here B + C = C. This implies B = 0; only if B = 0 will B + C be equal to C. But the addition of the first two digits that we carried out, gave us A; C + B = A. If B = 0, then C + B should be C, and not A.

This means that C + B is not A but is a double-digit number with A being its rightmost digit. This means C + B has a carry; the maximum carry can be one because even if C and B are both 9 (the highest single digit possible), yet the carry over will be one (9 + 9 = 18).

Therefore we conclude that C + B = 1A (1A can be 11, 12, 13, 14, 15, 16, 17, 18 or 19)

Therefore, C + B = 10 + A (suppose C + B = 18, then C + B = 10 + 8).

Now, we add the two middle digits taking the carry from the first addition. We have concluded that C + B has a carry, therefore the addition of the two middle digits will also have a carry. Thus 1 + B + C = 1C, where 1C can be 11, 12, 13, 14, 15, 16, 17, 18 or 19. That is: 1 + B + C = 10 + C.

Now we are getting somewhere. We have 1 + B + C = 10 + C. Therefore, B + C = 9 + C. Thus B = 9.

What remains is to add the leftmost digits, together with the carry. So we get 1 + A + A = B.

But B = 9; therefore, 1 + 2A = 9. So, 2A = 8, or A = 4.

We have C + B = 10 + A; therefore C + 9 = 10 + 4. Thus, C = 5.

So, A = 4, B = 9 and C = 5. The two numbers we then get are: ABC - 495 and ACB - 459; 495 (ABC) + 459 (ACB) = 954 (BCA).

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