elementary number theory - Prove that, $(2cdot 4 cdot 6 cdot ... cdot 4000)-(1cdot 3 cdot 5 cdot ...cdot 3999)$ 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 $2001$?

Answer

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 $$2\cdot4\cdot6\cdot...\cdot4000 - 1\cdot3\cdot5\cdot...\cdot3999 = 2001k_1 - 2001k_2 = 2001(k_1-k_2)$$