MIT Mathematical Tournament (IV)

We note that the product

$\begin{aligned} P & = 2\times 6 \times 10 \times \cdots \times (4n-2) \\ & = \prod_{k=1}^n (4k-2) = \prod_{k=1}^n 2(2n-1) = 2^n(2n-1)!! = \frac {2^n(2n)!}{2^nn!} = \frac {(2n)!}{n!} \end{aligned}$

Therefore, $\dfrac P{n!} = \dfrac {(2n)!}{n!n!} = \dbinom {2n}n$ which is an integer for all positive integers $n$ . Hence the answer is $\boxed{2005}$ .

Note that $2 \times 6 \times 10 \times \cdots \times (4n-2) = 2^n \times (2n-1)!! = 2^n \times \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{n!}$

Therefore $\frac{2 \times 6 \times 10 \times \cdots \times (4n-2)}{n!} = \frac{(2n)!}{n! \times n!} = \binom{2n}{n}$ is always an integer, so the answer is the number of integers between $1$ and $2005$ inclusive, or $\boxed{2005}$

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Fun fact: A stronger statement is also true: The product is actually divisible by $(n+1)!$ for all non-negative $n$ , as we can show $C_n := \frac{1}{n+1} \binom{2n}{n} = \binom{2n}{n} - \binom{2n}{n+1} \in \mathbb{Z}$