Example 8 Find the last 3 digits of $1 \times 3 \times 5 \times 7 \times \cdots \times 2005$.
Solution: (1) Find $a(0 \leqslant a \leqslant 9)$, such that $3^{2006} \equiv$ $a(\bmod 10)$.
$\because 3^{2} \equiv 9 \equiv-1(\bmod 10)$,
$\therefore 3^{4} \equiv 1(\bmod 10), 3^{2006} \equiv 3^{2004+2} \equiv 3^{4 \times 501+2}$ $\equiv 3^{2}(\bmod 10)$,
Therefore, the last digit of $3^{2006}$ is 9;
(2) Find $b(0 \leqslant b \leqslant 99)$, such that $3^{2006} \equiv$ $b(\bmod 100)$.

Notice: $4 \times 25=100$ and $(4,25)=1,3^{4} \equiv 81$ $\equiv 1(\bmod 5)$,
$\because 3^{4} \equiv 81 \equiv 6(\bmod 25), 3^{8} \equiv 36 \equiv 11(\bmod 25)$,
$\therefore 3^{12} \equiv 66 \equiv-9(\bmod 25), 3^{16} \equiv-54 \equiv$ $-4(\bmod 25)$,
$$
\begin{array}{l}
\therefore 3^{20} \equiv-24 \equiv 1(\bmod 25), \quad(1) \\
\because 3^{2} \equiv 1(\bmod 4), \therefore 3^{20} \equiv 1(\bmod 4),
\end{array}
$$

From (1) and (2), we get $3^{20} \equiv 1(\bmod 100)$,
$\therefore 3^{2006} \equiv 3^{20 \times 100} \cdot 3^{6} \equiv 29(\bmod 100)$, so the last two digits of $3^{2006}$ are 29.
Solution: Notice: $8 \times 125=1000$ and $(8,125)=1$,
$\because(2 n-3)(2 n-1)(2 n+1)(2 n+3) \equiv$ $\left(4 n^{2}-9\right)\left(4 n^{2}-1\right) \equiv 1(\bmod 8)$,

and $M=1 \times 3 \times 5 \times 7 \times \cdots \times 2005=125 m$ is the product of 1003 odd numbers,
$\therefore M \equiv 1 \times 3 \times 5 \equiv 7(\bmod 8), 125 m \equiv 5 m \equiv$ $7(\bmod 8), \therefore m \equiv 3(\bmod 8)$,
$\therefore M \equiv 125 m \equiv 125 \times 3 \equiv 375(\bmod 8)$,
$\therefore M \equiv 125 m \equiv 375(\bmod 8)$.
That is, the last 3 digits of $1 \times 3 \times 5 \times 7 \times \cdots \times 2005$ are 375.