Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$ Prove that the difference between the product of the first 2000 even numbers and the first $2000$ odd numbers is a multiple of $2001$. Please show the method.
I have started with the following process:
$$(2\cdot 4 \cdot 6 \cdots 4000)-(1\cdot 3 \cdot 5 \cdots 3999)$$ 
How we can proceed it to find that it is a multiple of $2001$?
 A: Hint: show that $2001$ is a factor of both products. One is very easy. For the other you need to know that $2001$ is not prime (you should be able to spot an easy factor).
A: We have $$2001=3\times 23 \times 29.$$ 
Now note that $$\begin{align}2\cdot 4 \cdot 6 \cdot ... \cdot 4000&=(2\times 1)\cdot (2 \times 2) \cdot (2 \times 3) \cdot ... \cdot  (2\times 2000)\\
&=2^{2000}\times 1\cdot2\cdot3\cdot... \cdot23 \cdot ... \cdot 29 \cdot....\cdot2000
\end{align}$$
Therefore $2\cdot 4 \cdot 6 \cdot ... \cdot 4000$ is a multiple of $2001$. Now since the other number is the multiplication of odd numbers including $3$, $23$, and $29$, it is also a multiple of $2001$. Therefore the difference is also a multiple of $2001$.
A: Try proving that it is equal to $2001k_1 - 2001k_2$.
The product of odds has $2001$ as its factor, hence it can be written as $2001k_2$. Now the product of evens has $667$ and $3$ as its factors and thus making $2001$ as its factor.
So $$2\cdot4\cdot6\cdot...\cdot4000 - 1\cdot3\cdot5\cdot...\cdot3999 = 2001k_1 - 2001k_2 = 2001(k_1-k_2)$$
