A5. Jan has a wooden cube. Each of the faces he divides into a $2 \times 2$ grid of squares, which he then paints in a black-and-white pattern: two diagonally adjacent squares are black, and the other two are white. At each corner of the cube, three squares come together. If two or three of these are black, we call the corner dark.

![](https://cdn.mathpix.com/cropped/2024_04_17_998065f2d90ec501d25cg-2.jpg?height=235&width=235&top_left_y=485&top_left_x=1590)

What is the smallest possible number of dark corners that the cube can have?
A) 0
B) 1
C) 2
D) 3
E) 4
A5. C) 2 On the cube, after distribution, there are a total of 24 squares: 12 white and 12 black. The number of dark vertices cannot be zero, because then the number of black squares would not exceed \(8 \times 1=8\) (one per vertex). The number of dark vertices also cannot be one, because then the number of black squares would not exceed \(3+7 \times 1=10\). In the figure, you can see a solution where only two vertices are dark. Therefore, the minimum number is 2.

![](https://cdn.mathpix.com/cropped/2024_04_17_20b3a516711f91b2fa07g-1.jpg?height=391&width=357&top_left_y=2346&top_left_x=1478)