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Revision as of 10:56, 4 July 2013

Problem

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

$\mathrm{(A)}\ 7 \qquad \mathrm{(B)}\ 41 \qquad \mathrm{(C)}\ 49 \qquad \mathrm{(D)}\ 289 \qquad \mathrm{(E)}\ 2009$

Solution

The prime factorization of $2009$ is $2009 = 7\cdot 7\cdot 41$. As $a<b<2009$, the ratio must be positive and larger than $1$, hence there is only one possibility: the ratio must be $7$, and then $b=7\cdot 41$, and $a=41\Rightarrow\text{(B)}$.

See Also

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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