Example 2. Find the remainder when $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$ is divided by
81.
Solve $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$
$$
\begin{array}{l}
=\left(2^{22}-2^{20}+2^{18}-2^{16}+2^{14}-\right. \\
\left.2^{12}\right)+\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
=\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \\
\left(2^{12}+1\right)
\end{array}
$$

Since $9=(1001)_{2}, 81=9 \times 9$,
and $2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1$
$$
\begin{aligned}
= & (10001000100)_{2}-(100010001)_{2} \\
= & (1100110011)_{2} \\
= & (1001)_{2}(1011011)_{2} \\
& 2^{12}+1=(\underbrace{11 \cdots 1)_{2}}_{12 \text { 个 }}+(10)_{2} \\
= & (1001)_{2} \cdot(111000111)_{2}+(10)_{2}
\end{aligned}
$$

Therefore, $\left(2^{10}-2^{8}+2^{6}-2^{4}+2^{2}-1\right) \cdot\left(2^{12}\right.$ $+1)$ modulo 81 is the same as $(1001)_{2}(1011011)_{2}$ $\cdot(10)_{2}$ modulo $(1001)_{2} \cdot(1001)_{2}$, which is $(1011011)_{2} \cdot(10)_{2}$ modulo $(1001)_{2}$.
And $(1011011)_{2} \cdot(10)_{2}=(10110110)_{2}$ $=(1001)_{2} \cdot(10100)_{2}+(10)_{2}$,
thus the remainder of $2^{22}-2^{20}+2^{18}-\cdots+2^{2}-1$ divided by 81 is $(10)_{2}=2$.