Find the remainder when $10^{2^{0}}+10^{2^{1}}+\cdots+10^{2^{2021}}$ is divided by $\underbrace{44 \cdots 44}_{44}$.
Let $N=10^{2^{0}}+10^{2^{1}}+\cdots+10^{2^{2021}}$. By CRT, it suffices to find $N$ $\bmod 4$ and $\bmod M=\underbrace{11 \cdots 11}_{44}$.

Modulo 4, the remainder is 2. Modulo $M=\left(10^{44}-1\right) / 9,10^{44}=1$, so $10^{2^{i}}(\bmod M)$ is determined by $2^{i}(\bmod 44)$.

Mod $44,2^{0}$ is 1 and $2^{1}$ is 2 . The remaining powers of two are all divisible by 4 . Furthermore, the period of $2(\bmod 11)$ is 10 , so the remaining 2020 powers of 2 are in fact divided equally among 10 nonzero remainders that are multiples of 4 . Therefore, $N(\bmod M)$ is

$$
10^{1}+10^{2}+\frac{2020}{10} \cdot \sum_{i=1}^{10} 10^{4 i}=2020202020202020202020202020202020202020110
$$

which is $2(\bmod 4)$, and is therefore the answer.