SolUTION
C
Using the expansion $(x+y)^{2}=x^{2}+2 x y+y^{2}$, we have
$$
\begin{aligned}
\left(10^{2020}+2020\right)^{2} & =\left(10^{2020}\right)^{2}+2 \times 10^{2020} \times 2020+2020^{2} \\
& =10^{4040}+4040 \times 10^{2020}+4080400 .
\end{aligned}
$$

We now note that when we add the three terms $10^{4040}, 4040 \times 10^{2002}$ and 408040 , no two non-zero digits occur in the same column:
$$
\begin{array}{l}
100 \ldots \ldots \ldots . .0000 \ldots \ldots . . . . . . . . .0000000 \\
+\quad 4040 \ldots \ldots . . . . .0000000 \\
+\frac{4080400}{100 \ldots \ldots \ldots .4040 \ldots \ldots \ldots .4080400} \\
\end{array}
$$

It follows that the non-zero digits in the final answer are exactly the non-zero digits in the three terms in (1) above.
Therefore the sum of the digits in the number $\left(10^{2020}+2020\right)^{2}$ is
$$
1+4+4+4+8+4=25 \text {. }
$$

Commentary
The exact number of 0 s in the final answer is not important. What matters is that the non-zero digits in the three terms $10^{4040}, 4040 \times 10^{2020}$ and 4080400 are in different columns, and so, when they are added, the non-zero digits in the final answer are exactly the non-zero digits in these terms. However, you are asked to work out the number of 0 s in Problem 21.1 below.

FOR INVESTIGATION
21.1 Find the positive integers $m$ and $n$ so that
$$
\left(10^{2020}+2020\right)^{2}=1 \overbrace{000 \ldots 000}^{m} 404 \overbrace{000 \ldots 000}^{n} 4080400 .
$$
21. When written out in full, the number $\left(10^{2020}+2020\right)^{2}$ has 4041 digits.
What is the sum of the digits of this 4041-digit number?
A 9
B 17
C 25
D 2048
E 4041