4. Vasya drew a grid figure. It turned out that it can be cut into $2 \times 2$ squares, and it can also be cut into zigzags of 4 cells. How many cells can be in Vasya's figure?

## Solution:
Answer: any number that is a multiple of 8, starting from 16.

Example:

![](https://cdn.mathpix.com/cropped/2024_05_06_d99ea33c8d1f8869f2c9g-1.jpg?height=308&width=394&top_left_y=2293&top_left_x=840)

Evaluation: we will prove that the number of cells is a multiple of 8, that is, that the number of figures obtained by cutting is even. This can be done in two ways.

Method 1. We will color the board in a checkerboard pattern of $2 \times 2$. Each square occupies an even number of cells, and each zigzag occupies an odd number. Therefore, the number of figures is even.

Method 1'. We will color the board in a staircase pattern. Each square occupies an odd number of cells, and each zigzag occupies an even number. Therefore, the number of figures is even.
![](https://cdn.mathpix.com/cropped/2024_05_06_d99ea33c8d1f8869f2c9g-2.jpg?height=320&width=970&top_left_y=591&top_left_x=540)

On the left is the image for Method 1, on the right - for $1^{\prime}$.

Method 2. Notice that in each row and each column of such figures, there is an even number of cells, as it can be cut into $2 \times 2$ squares.

Cut the figure into diagonal tetrominoes and look at the vertical ones. In the topmost row, there is an even number of vertical diagonal tetrominoes. Remove them. From each row and each column, an even number of cells has been removed. Repeat this operation sequentially from top to bottom for each row. In the end, we get that the number of vertical diagonal tetrominoes is even. Similarly, the number of horizontal tetrominoes is even.

It remains to note that 8 does not work. This is obvious, as we have only two different ways to form a figure from two squares.