2006 AMC 8 Problem 24

Attempt Problem 24 of the 2006 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 2006 AMC 8 solutions, or check the answer key.

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

24.

In the multiplication problem below A,A, B,B, C,C, DD are different digits. What is A+B?A+B? ABA×CDCDCD \begin{array}{cccc}& A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}
11
22
33
44
99
Answer: A
Concepts:cryptarithmplace value
Small hint:
Write CDCDCDCD as 100⋅CD+CD100\cdot CD+CD.
Big hint:
Since CDCD is nonzero, divide the equation by the two-digit number CDCD.
Solution:
Note that CDCD=101⋅CD. CDCD = 101 \cdot CD. This forces ABA=101.ABA = 101. Then A=1,A = 1, B=0,B = 0, and A+B=1.A + B = 1.
Thus, A is the correct answer.
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