1997 AMC 12 Problem 1

Attempt Problem 1 of the 1997 AMC 12 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 1997 AMC 12 solutions, or check the answer key.

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

1.

If aa and bb are digits for which 2a×b369920989 \begin{array}{r} 2a\\[-2pt] {}\times b3\\ \hline 69\\ 92\phantom0\\ \hline 989 \end{array} then a+b=a+b=
33
44
77
99
1212
Answer: C
Concepts:long multiplicationdigits
Small hint:
Use the first partial product to determine the two-digit top factor
Big hint:
The second partial product is the top factor multiplied by bb
Solution:
The first partial product says 3(20+a)=69,3(20+a)=69, so a=3.a=3. The second says 23b=92,23b=92, so b=4.b=4. Thus a+b=7,a+b=7, and the correct answer is C.
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