# Task № 8.1

## Condition:

A board of $2022 \times 2022$ is given. Olya and Masha take turns coloring $2 \times 2$ squares on it red and blue, with the sides of the squares aligned with the grid lines. The girls agreed that each cell can be painted no more than once in blue and no more than once in red. Cells that are painted blue and then red (or vice versa) become purple. Once all cells are painted, the girls counted how many of them are purple. What results could they have obtained?

## Answer options:

○ $2022 \cdot 2022$

○ $2022 \cdot 2021$

○ $2022 \cdot 2020$

○ $2020 \cdot 2020$

## Correct answers:

○ $2022 \cdot 2020$

○ $2020 \cdot 2020$

## Exact match of the answer -1 point
# Solution.

Let the number of purple cells be denoted by \( x \). Each move colors a \( 2 \times 2 \) square, meaning the number of "colorings" is \( 4k \), where \( k \) is the number of moves. On the other hand, let \( x \) be the number of purple cells, which are the cells that the girls painted twice, so \( 2x \) "colorings" were used on them. Then the number of cells painted once is \( 2022 \times 2022 - x \). Therefore, the total number of "colorings" is \( 2022 \times 2022 - x + 2x = 2022 \times 2022 + x \). Thus, we have calculated the number of "colorings" in two ways: \( 2022 \times 2022 + x = 4k \).

From this, \( x = 4(k - 1011 \times 1011) \). Therefore, \( x \) is divisible by 4. Thus, the options \( 2022 \times 2021 \) and \( 2021 \times 2021 \) are ruled out. Moreover, we will prove that all cells cannot be purple. If this were the case, each cell would have been painted twice. Then consider the last colored square. All its cells were already colored before, meaning the entire board was already colored, so the girls would have stopped painting at that moment. Thus, the option \( 2022 \times 2022 \) is ruled out. Note that any other number of the form \( 4k \) is possible. Divide the board into squares, a total of \( 1011 \times 1011 \). Then Masha can repeat Olya's actions \( k \) times, so \( k \) squares will become purple. Then they will take turns painting the remaining \( 1011 \times 1011 - k \) (each one square at a time).