Example 2 Find the last two digits of $3^{406}$ when written as a decimal number.
Solve: This is to find the smallest non-negative remainder $b$ when $3^{106}$ is divided by 100, i.e., $b$ satisfies
$$3^{406} \equiv b(\bmod 100), \quad 0 \leqslant b \leqslant 99$$

Notice that $100=4 \cdot 25, (4,25)=1$. Clearly, $3^{2} \equiv 1(\bmod 4), 3^{4} \equiv 1(\bmod 5)$. Note that 4 is the smallest power, by Example 5 in Chapter 1, §4, we know that for $3^{d} \equiv 1(\bmod 25)$ to hold, $4 \mid d$ must be true. Therefore, calculate:
$$\begin{array}{c}
3^{4} \equiv 81 \equiv 6(\bmod 25), \quad 3^{8} \equiv 36 \equiv 11(\bmod 25) \\
3^{12} \equiv 66 \equiv-9(\bmod 25), \quad 3^{16} \equiv-54 \equiv-4(\bmod 25) \\
3^{20} \equiv-24 \equiv 1(\bmod 25)
\end{array}$$

From this and $3^{20} \equiv 1(\bmod 4)$, by Property IX, we derive $3^{20} \equiv 1(\bmod 100), 3^{400} \equiv 1(\bmod 100)$. Therefore, $3^{406} \equiv 3^{400} \cdot 3^{6} \equiv 3^{6} \equiv 29(\bmod 100)$. So, the unit digit is 9, and the tens digit is 2.

If we do not use the property $4 \mid d$, we have to calculate the remainder $b_{j}$ of $3^{j}$ modulo 25 (to facilitate the calculation of $b_{j}$, the absolute smallest remainder should be taken), as shown in the table below:
\begin{tabular}{|c|cccccccccc|}
\hline$j$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$b_{j}$ & 3 & 9 & 2 & 6 & -7 & 4 & 12 & 11 & 8 & -1 \\
\hline
\end{tabular}

From the $3^{10} \equiv-1(\bmod 25)$ obtained here, we conclude $3^{20} \equiv 1(\bmod 25)$.