2009 AMC 10A Problem 9

Attempt Problem 9 of the 2009 AMC 10A 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 2009 AMC 10A solutions, or check the answer key.

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

9.

Positive integers a,a, b,b, and 2009,2009, with a<b<2009,a \lt b \lt 2009, form a geometric sequence with an integer ratio. What is a?a?
77
4141
4949
287287
20092009
Answer: B
Concepts:geometric sequenceprime factorization
Small hint:
If the ratio is r,r, then ar2=2009a r^2 = 2009
Big hint:
Factor 2009=72⋅412009 = 7^2 \cdot 41 to find the only integer ratio greater than 11
Solution:
Let the common ratio be r.r. Then ar2=2009=72⋅41.a r^2 = 2009 = 7^2 \cdot 41.
Since rr must be an integer greater than 1,1, the only possibility is r=7,r = 7, giving a=41a = 41 and the sequence 41,287,2009.41, 287, 2009.
Thus, the correct answer is B.
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