## Task A-4.5.

Determine the last three digits of the number $21^{2021}$.
## First Solution.

Using the binomial theorem, we get

$$
\begin{array}{rlr}
21^{2021} & =(1+20)^{2021}=1+2021 \cdot 20+\binom{2021}{2} \cdot 20^{2}+\binom{2021}{3} \cdot 20^{3}+\cdots & 2 \text { points } \\
& =1+40420+2021 \cdot 1010 \cdot 400+\binom{2021}{3} \cdot 8000+\cdots & \\
1 \text { point }
\end{array}
$$

$$
=40421+1000 N
$$

for some natural number $N$ (the unlisted terms contain a factor of $20^{k}$, where $k \geqslant 4$, and are therefore divisible by 1000).

The last three digits of the number $21^{2021}$ are thus 421.

1 point