Problem (Number Theory):
Compute the integer $n$ such that $2009<n<3009$ and the sum of the odd positive divisors of $n$ is 1024 .

Solution:
Suppose that $n=2^{k} p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$, where the $p_{i}$ are distinct odd primes, $k$ is a nonnegative integer, and $a_{1}, \ldots, a_{r}$ are positive integers. Then the sum of the odd positive divisors of $n$ is equal to

$$
\prod_{i=1}^{r}\left(1+p_{i}+\cdots+p_{i}^{a_{i}}\right)=\prod_{i=1}^{r} \frac{p_{i}^{a_{i}+1}-1}{p_{i}-1}=1024=2^{10}
$$

Note that $1+p_{i}+\cdots+p_{i}^{a_{i}}$ is the sum of $a_{i}+1$ odd numbers. Because the product of those sums is a power of two, each sum must be even (in fact, a power of 2). Thus, each $a_{i}$ must be odd.

Because $1+11+11^{2}+11^{3}>1024$, if $p_{i} \geq 11$, then $a_{i}=1$ and $1+p_{i}$ must be a power of 2 that is no greater than 1024. The possible values of $p_{i}$, with $p_{i} \geq 11$, are 31 and 127 (as 5 divides 255, 7 divides 511, and 3 divides 1023).

If $p_{1}<11$, then $p_{i}$ can be $3,5,7$. It is routine to check that $a_{i}=1$ and $p_{i}=3$ or 7 .

Thus $a_{i}=1$ for all $i$, and the possible values of $p_{i}$ are $3,7,31,127$. The only combinations of these primes that yield 1024 are $(1+3) \cdot(1+7) \cdot(1+31)\left(\right.$ with $\left.n=2^{k} \cdot 3 \cdot 7 \cdot 31=651 \cdot 2^{k}\right)$ and $(1+7) \cdot(1+127)$ (with $n=7 \cdot 127=889 \cdot 2^{k}$ ). Thus $n=651 \cdot 2^{2}=\mathbf{2 6 0 4}$ is the unique value of $n$ satisfying the conditions of the problem.

Final answer: 2604