problem: Positive integers $a$, $b$, and $2009$, with $a<b<2009$, form a geometric sequence with an integer ratio. What is $a$?
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 thus $b=7\cdot 41$ and $a=\boxed{41}$.