How many positive integers satisfy the double inequality $2000<\sqrt{n(n-1)}<2005$?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
The correct option is (e).

Since the numbers that appear are all positive, we can square them while maintaining the direction of the inequalities, obtaining

$$
2000 \times 2000=2000^{2}<n(n+1)<2005^{2}=2005 \times 2005
$$

Notice that $n$ and $n+1$ are consecutive integers, so the only options are the following.

- $2000^{2}<2000 \times 2001<2005^{2}$
- $2000^{2}<2001 \times 2002<2005^{2}$
- $2000^{2}<2002 \times 2003<2005^{2}$
- $2000^{2}<2003 \times 2004<2005^{2}$
- $2000^{2}<2004 \times 2005<2005^{2}$

Therefore, we have five possibilities for $n$, namely, 2000, 2001, 2002, 2003, and 2004.