Prove that, $(2\cdot 4 \cdot 6 \cdot ... \cdot 4000)-(1\cdot 3 \cdot 5 \cdot ...\cdot 3999)$ is a multiple of $2001$

Question

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 1

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

Answer 2

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).

Answer 3

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 $2001$.