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

## Problem

Positive integers $a$, $b$, and $2009$, with $a, 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, 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 $a=41\Rightarrow\text{(B)}$.