1991 AMC 8 Problem 20

Attempt Problem 20 of the 1991 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

In the addition problem shown, each digit has been replaced by a letter. If different letters represent different digits, then C=C =
ABCAB+A300\begin{array}{cccc} & A & B & C \\ & & A & B \\ + & & & A \\ \hline & 3 & 0 & 0 \end{array}
11
33
55
77
99
Answer: A
Concepts:cryptarithmplace value
Small hint:
Write the sum in terms of A,B,CA, B, C
Big hint:
100A+10B+C100A + 10B + C +10A+B+A+ 10A + B + A =111A+11B+C= 111A + 11B + C =300= 300
Solution:
The three numbers add to 111A+11B+C=300.111A + 11B + C = 300. Since A=1A = 1 is too small and A≥3A \ge 3 is too large, A=2.A = 2.
Then 11B+C=78,11B + C = 78, which forces B=7B = 7 and C=1.C = 1. So C=1.C = 1.
Thus, the correct answer is A .
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