[Number Theory]
Problem: If $n$ is a positive integer, then $n$ !! is defined to be $n(n-2)(n-4) \cdots 2$ if $n$ is even and $n(n-2)(n-4) \cdots 1$ if $n$ is odd. For example, $8 ! !=8 \cdot 6 \cdot 4 \cdot 2=384$ and $9 ! !=9 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945$. Compute the number of positive integers $n$ such that $n !$ ! divides 2012!!.

Solution: If $n$ is even and $n \leq 2012$, then $n$ !! $\mid 2012$ !! trivially, while if $n>2012,2012$ !! $<n$ !!, so $n$ !! cannot divide 2012!!. Thus there are a total of 1006 even values of $n$ such that $n$ !! | 2012!!. If $n$ is odd and $n<1006$, then $n$ !! | 2012!!. To show this, rearrange the terms of 2012!! and factor:

$$
\begin{aligned}
2012 ! ! & =2 \cdot 4 \cdot 6 \cdots 2010 \cdot 2012 \\
& =(2 \cdot 6 \cdot 10 \cdots 2010)(4 \cdot 8 \cdot 12 \cdots 2012) \\
& =2^{503}(1 \cdot 3 \cdot 5 \cdots 1005)(4 \cdot 8 \cdot 12 \cdots 2012)
\end{aligned}
$$

However, the condition $n<1006$ is not necessary, only sufficient, because $n$ !! also divides 2012 if $1007 \cdot 1009 \cdots n \mid(4 \cdot 8 \cdot 12 \cdots 2012)$. (The factor of $2^{503}$ is irrelevant because all the factors on the left side are odd.) The expression $(4 \cdot 8 \cdot 12 \cdots 2012)$ can be factored as $4^{503}(1 \cdot 2 \cdot 3 \cdot \cdots 503)=4^{503} \cdot 503$ !. Examining the numbers $1007,1009, \ldots$ in sequence shows that 1007 is satisfactory, because $1007=19 \cdot 53$. On the other hand, 1009 is prime, so it cannot be a factor of $4^{503} \cdot 503$ !. Thus the largest possible odd value of $n$ is 1007 , and there are 504 odd values of $n$ altogether. The total is $1006+504=\mathbf{1 5 1 0}$.

Final answer: 1510