2. In the cells of an $11 \times 11$ square, zeros and ones are arranged in such a way that in any figure of four cells in the shape of $\square$, the sum of the numbers is odd. (The figure can be rotated and flipped). What is the minimum number of ones that can be in such an arrangement?
Answer: 25

Solution. Place 25 figures in the square without any common cells (see the left figure). Each of them contains at least one unit, so the total number of units is no less than 25. A suitable arrangement of 25 units: in all cells with even coordinates (see the right figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-01.jpg?height=502&width=505&top_left_y=1825&top_left_x=547)

![](https://cdn.mathpix.com/cropped/2024_05_06_2ae48c21b42f492bf703g-01.jpg?height=494&width=489&top_left_y=1829&top_left_x=1109)