If AA + BB + CC = ABC, then what is the value of A + B + C?

Concept: A repeated two-digit number like AA stands for 11 x A (since AA = 10A + A = 11A). To find the unknown digits in a cryptarithmetic puzzle, rewrite every term in place-value form, add the units, tens, and hundreds columns separately (carrying over as needed), and solve the resulting digit equation - you are matching digits, not doing an ordinary sum.

Applying this to the puzzle AA + BB + CC = ABC:

1. Since AA = 11A, BB = 11B, and CC = 11C, the puzzle becomes 11A + 11B + 11C = 100A + 10B + C (because ABC, read as a number, equals 100A + 10B + C).

2. Simplifying: 11A + 11B + 11C - 100A - 10B - C = 0, which reduces to 89A = B + 10C.

3. A is the leading digit of the 3-digit sum ABC, so A is at least 1; B and C are single digits (0-9), so B + 10C can be at most 99. Since 89A must be at most 99, the only possible value is A = 1 (A = 2 would need 89 x 2 = 178, which is impossible).

4. With A = 1: 89 = B + 10C. Testing values of C from 0-9, only C = 8 gives a valid single digit for B: B = 89 - 80 = 9.

5. So A = 1, B = 9, C = 8.

6. A + B + C = 1 + 9 + 8 = 18.

Cross-check by direct addition: AA + BB + CC = 11 + 99 + 88 = 198, and ABC with A = 1, B = 9, C = 8 is indeed 198 - the two sides match, confirming the digits.

So A + B + C = 18.