What are the last three digits of the product of the odd numbers from 1 to 1000?

Question

What are the last three digits of the product of the odd numbers from $1$ to $1000$?

Thank you in advance, any help is appreciated.

Answer 1

We have that $999!!$ is a multiple of $125$ since $125$ is one of its factors.
On the other hand, $999\equiv 7\pmod{8}$, hence

$$ 999!!=\prod_{n=0}^{499}(2n+1)\equiv(1\cdot 3\cdot 5\cdot 7)^{125}\equiv 1\pmod{8} $$ By the Chinese remainder theorem, it follows that $$ 999!!\equiv \color{red}{625}\pmod{1000}.$$