equation $\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}=4^x $ How to solve the equation:
$$\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}=4^x $$
 A: $$\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}$$
$$=\frac{(2008\cdot2006\cdot2004\cdots1006)(2\cdot4\cdot6\cdots1004)}{(1\cdot3\cdot5\cdots1003)(2\cdot4\cdot6\cdots1004)}$$
$$=\frac{2\cdot4\cdot6\cdots 2006\cdot2008}{(1004)!}$$
$$=\frac{\prod_{1\leq r\leq 1004} (2\cdot r)}{(1004)!}$$
$$=2^{1004}\frac{\prod_{1\leq r\leq 1004}r}{(1004)!}$$
$$=2^{1004}=4^{502}$$
$$\implies x=502$$
More generally, $$\frac{4n(4n-2)(4n-4)\cdots(2n+2)}{1\cdot3\cdot5\cdots(2n-1)}$$
$$=\frac{4n(4n-2)(4n-4)\cdots(2n+2)(2\cdot4\cdots 2n)}{1\cdot3\cdot5\cdots(2n-1)(2\cdot4\cdots 2n)}$$
$$=\frac{\prod_{1\leq r\leq 2n} (2\cdot r)}{(2n)!}$$
$$=2^{2n}$$  
Here evidently, $n=502$.
A: Hint:  If you trust the equation to have an integral solution, all the odd numbers will have to divide out.  I suspect it does, but you could chase them around to prove it.  There are no factors of $2$ in the denominator, so all you have to do is count the factors in the numerator.  Every number has at least one, every second has at least two, every fourth has three, etc.
