3. What is the sum of the digits of the number $20^{21}+\left(10^{2021}+21\right)^{2}$?
(A) 21
(B) 30
(C) 37
(D) 42
(E) A number greater than 100.
3. The answer is $(\mathbf{D})$. We can expand the square

$$
N=20^{21}+\left(10^{2021}+21\right)^{2}=10^{4042}+42 \cdot 10^{2021}+2^{21} \cdot 10^{21}+21^{2}
$$

We observe that there are no carries between the written addends, so the result is the sum of the digit sums of $1, 42, 2^{21}, 21^{2}$.

Since $2^{21}=2 \cdot 1024^{2}=2097152$ has a digit sum of 26 and $21^{2}=441$ has a digit sum of 9, the solution is $1+6+26+9=42$.