# equation $\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}=4^x$

How to solve the equation: $$\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}=4^x$$

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What have you tried so far? –  Michael Albanese Sep 22 '12 at 13:54
If you mean $$\frac{2008\cdot 2006\cdot 2004\cdots 1006}{1\cdot 3\cdot 5\cdots 1003}=4^x ,$$ replace $.$ with $\cdot$ (\cdot) and $...$ with $\cdots$ (\cdots). –  Américo Tavares Sep 22 '12 at 13:59

$$\frac{2008\cdot2006\cdot2004\cdots1006}{1\cdot3\cdot5\cdots1003}$$

$$=\frac{(2008\cdot2006\cdot2004\cdots1006)(2\cdot4\cdot6\cdots1004)}{(1\cdot3\cdot5\cdots1003)(2\cdot4\cdot6\cdots1004)}$$

$$=\frac{2\cdot4\cdot6\cdots 2006\cdot2008}{(1004)!}$$

$$=\frac{\prod_{1\leq r\leq 1004} (2\cdot r)}{(1004)!}$$

$$=2^{1004}\frac{\prod_{1\leq r\leq 1004}r}{(1004)!}$$

$$=2^{1004}=4^{502}$$

$$\implies x=502$$

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, $n=502$.

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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.

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