2009 AMC 10A Problem 9

1. Call the whole-number ratio $r$. The terms are $a$, then $a \times r$, then $a \times r \times r$. Since the terms get bigger, $r$ is at least $2$.
2. The third term is $2009$, so
$$a \times r^2 = 2009.$$
3. Break $2009$ into primes: $2009 = 7 \times 287 = 7 \times 7 \times 41$.
4. So $r^2$ must be a perfect square that divides $7 \times 7 \times 41$. The only such square bigger than $1$ is $7 \times 7 = 49$. So $r = 7$.
5. Then $a = 2009 \div 49 = 41$.
6. Check: $41$, $41 \times 7 = 287$, $287 \times 7 = 2009$. The terms increase and are whole numbers.

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

With a whole-number ratio, the third term is the first term times a perfect square. So the question becomes "which perfect squares divide $2009$?", and the prime factorization answers it right away. The year of the contest shows up often on the AMC, so it pays to know how it factors.

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 $\boxed{\textbf{(B)}\ 41}$.