Example 5 Find the last three digits of $1 \times 3 \times 5 \times \cdots \times 1997$.
Solution: According to the problem, we only need to find the remainder of $M=1 \times 3 \times 5 \times \cdots \times 1997$ modulo 1000.

Obviously, $125 \mid M$, i.e., $M=125 m$, where $m$ is an integer, so we only need to consider the remainder modulo 8.

Since $(2 n-3) \cdot(2 n-1) \cdot(2 n+1) \cdot(2 n+3) \equiv 1(\bmod 8)$, for any integer $n$, then by $M$ being the product of 999 consecutive odd numbers and $999=4 \times 249+3$, we get $M \equiv 1 \times 3 \times 5 \equiv 7(\bmod 8)$, i.e., $125 m \equiv 7(\bmod 8)$, so $5 m \equiv 7(\bmod 8)$, which means $m \equiv 3(\bmod 8)$,

Therefore, $M \equiv 3 \times 125(\bmod 1000)$, i.e., $M \equiv 375(\bmod 1000)$.