32. Find the remainder when the integer
$$
1 \times 3 \times 5 \times 7 \times \cdots \times 2003 \times 2005
$$
is divided by 1000 .
32. Ans: 375
Let $N=1 \times 3 \times \cdots \times 2005$. We need to find the remainder when $N$ is divided by 1000 . Observe that $1000=8 \times 125$. Let $M=\frac{N}{125}$. We are going to find the remainder when $M$ is divided by 8 . Observe that
$$
\begin{aligned}
(2 n-3)(2 n-1)(2 n+1)(2 n+3) & =\left(4 n^{2}-1\right)\left(4 n^{2}-9\right) \\
& =16 n^{4}-40 n^{2}+9 \\
& \equiv 1(\bmod 8) .
\end{aligned}
$$

Thus the product $P$ of any 4 consecutive odd integers satisfies $P \equiv 1(\bmod 8)$. Write
$$
M=(1 \times 3 \times \cdots \times 123) \times(127 \times 129 \times \cdots \times 2005) .
$$

There are 62 factors in the expression $1 \times 3 \times \cdots \times 123$. Thus,
$$
\begin{aligned}
1 \times 3 \times 5 \times \cdots \times 123 & =1 \times 3 \times(5 \times \cdots \times 123) \\
& \equiv 3 \times 1^{15}(\bmod 8) \\
& \equiv 3(\bmod 8) .
\end{aligned}
$$

Similarly, there are 940 factors in the expression $127 \times \cdots \times 2005$. Thus,
$$
\begin{aligned}
127 \times 129 \times \cdots \times 2005 & \equiv 1^{235}(\bmod 8) \\
& \equiv 1(\bmod 8) .
\end{aligned}
$$

Hence,
$$
M \equiv 1 \times 3 \equiv 3(\bmod 8) .
$$

In other words,
$$
M=8 n+3
$$
for some positive integer $n$. Now, we have
$$
\begin{aligned}
N & =125 \times M \\
& =125 \times(8 n+3) \\
& =1000 n+375 .
\end{aligned}
$$

Therefore, the remainder is 375 .