2009 AMC 10A Problem 9

Let the common ratio be the integer $r$. Since the terms increase, $r \ge 2$. The terms are $a$, $ar$, $ar^2$, and the third equals $2009$:
$$a r^2 = 2009.$$

Factor: $2009 = 7 \cdot 287 = 7 \cdot 7 \cdot 41 = 7^2 \cdot 41$.

For $ar^2 = 2009$ with $r \ge 2$, the perfect square $r^2$ must divide $7^2 \cdot 41$. The only square divisor greater than $1$ is $49$, so $r = 7$ and
$$a = \frac{2009}{49} = 41.$$

The sequence is $41, 287, 2009$, which checks.

The answer is $\boxed{\textbf{(B)}\ 41}$.

A geometric sequence with integer ratio makes the last term a multiple of $r^2$, so the question is really "which perfect squares divide $2009$?" Prime factorization answers that instantly. Knowing the factorization of the contest year is a recurring time-saver on AMC problems.

• Reporting the square factor 49 as a instead of the cofactor 41, or missing the factorization 2009 = 7 7 41.
• Allowing $r = 1$ (which would violate $a < b < 2009$) and answering $2009$.