2009 AMC 10A Problem 9 — Solution | LIVE by Po-Shen Loh

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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.

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9.

Positive integers $a,$ $b,$ and $2009,$ with $a \lt b \lt 2009,$ form a geometric sequence with an integer ratio. What is $a?$

$7$

$41$

$49$

$287$

$2009$

Answer: B

Concepts:geometric sequence prime factorization

Difficulty rating: 1240

Solution:

Let the common ratio be $r.$ Then $a r^2 = 2009 = 7^2 \cdot 41.$

Since $r$ must be an integer greater than $1,$ the only possibility is $r = 7,$ giving $a = 41$ and the sequence $41, 287, 2009.$

Thus, the correct answer is B.

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