## Task A-1.1.

Which number is greater

$$
A=2022 \cdot\left(2024^{2}-2024+1\right) \quad \text { or } \quad B=2024^{3}-3 \cdot 2024^{2}+3 \cdot 2024 ?
$$
## Solution.

Let $n=2024$.

Subtract the observed numbers.

We have

$$
\begin{aligned}
B-A & =n^{3}-3 n^{2}+3 n-(n-2) \cdot\left(n^{2}-n+1\right) & & \\
& =n^{3}-3 n^{2}+3 n-\left(n^{3}-3 n^{2}+3 n-2\right) & & 2 \text { points } \\
& =2 . & & 1 \text { point }
\end{aligned}
$$

We conclude that $B$ is greater than the other two numbers.

Note: Solutions that do not achieve any other points in this task achieve 1 point if they introduce a notation like $n=2024$.

It is also possible to directly calculate the values for $A$ and $B$:

$$
\begin{aligned}
& A=2022 \cdot(4096576-2024+1)=8279186166 \\
& B=8291469824-3 \cdot 4096576+6072=8279186168
\end{aligned}
$$

Correct calculation of one of the numbers $A$ or $B$ earns 3 points, the remaining one 2 points, and the final conclusion 1 point.