8. Joep places the numbers 1 through 8 at the vertices of a cube: each vertex gets one number. For each face, he then adds up the numbers at the four vertices of that face and writes the result on the face. Then a net of the cube is made. Below are five figures given. Only one of them can be the net of Joep's cube.
![](https://cdn.mathpix.com/cropped/2024_04_17_01cb006263f00c7722c1g-2.jpg?height=178&width=1490&top_left_y=2296&top_left_x=290)

Which one is it?
A) the first
B) the second
C) the third
D) the fourth
E) the fifth

## B-questions
A8. E) the fifth We examine which figure could be the result of Joep's cube.

Two opposite faces of the cube together use all eight numbers exactly once. The numbers on opposite faces must therefore add up to $1+$ $2+\cdots+8=36$. The first figure is therefore eliminated, as it has the numbers 18 and 17 on opposite faces while $18+17 \neq 36$. The second figure is also eliminated, as it has 18 and 20 on opposite faces.

The third figure is eliminated because it is not a valid result of a cube: when we fold the figure, the squares with the numbers 10 and 19 overlap.

The number on each face must be at least $1+2+3+4=10$. This eliminates the fourth figure as well.

The fifth figure, finally, could be Joep's result. Below you can see how the numbers 1 to 8 can be distributed over the vertices such that the numbers at the four vertices of each face add up to 18.

![](https://cdn.mathpix.com/cropped/2024_04_17_fc21619166b069277c26g-3.jpg?height=417&width=397&top_left_y=914&top_left_x=835)