The vertices of a cube are numbered from 1 to 8, such that one of the faces has the vertices $\{1,2,6,7\}$ and the other five have the vertices $\{1,4,6,8\},\{1,2,5,8\}$, $\{2,3,5,7\},\{3,4,6,7\}$ and $\{3,4,5,8\}$. Which is the number of the vertex that is farthest from the vertex numbered 6?
(a) 1
(b) 3
(c) 4
(d) 5
(e) 7
The correct option is (d).

Solution 1: By drawing the cube and numbering its vertices according to the question statement, we obtain a figure in which we can see that vertex 5, being diametrically opposite, is the farthest from vertex 6.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-040.jpg?height=377&width=371&top_left_y=2364&top_left_x=1462)

Solution 2: Vertex 6 is on the faces $\{1,2,6,7\},\{1,4,6,8\}$, and $\{3,4,6,7\}$. Since the only vertex not appearing in these faces is 5, it follows that this is the vertex diagonally opposite to 6, that is, 5 is the vertex farthest from 6.