## Task A-1.1.

Determine the sum of all digits in the decimal representation of the number $\left(10^{2020}+2020\right)^{2}$.
## Solution.

By expanding the square of the sum we have

$\left(10^{2020}+2020\right)^{2}=\left(10^{2020}\right)^{2}+2 \cdot 10^{2020} \cdot 2020+2020^{2}=10^{4040}+404 \cdot 10^{2021}+4080400 . \quad 2$ points The number $10^{4040}$ consists of one unit followed by 4040 zeros. 1 point The number $404 \cdot 10^{2021}$ consists of the digits 4, 0, 4 followed by 2021 zeros. 1 point Therefore, the number $10^{4040}+404 \cdot 10^{2021}+4080400$ looks like

$$
1 \underbrace{00 \ldots 0}_{2016 \text { times }} 404 \underbrace{00 \ldots 0}_{2014 \text { times }} 4080400 .
$$

Thus, the sum of the digits is $1+4+4+4+8+4=25$.