Difference between revisions of "2009 AMC 10A Problems/Problem 9"

Revision as of 11:56, 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 $a=41\Rightarrow\text{(B)}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2009 AMC 10A Problems/Problem 9"

Revision as of 11:56, 4 July 2013

Problem

Solution

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