The eight vertices of a cube are randomly labelled with the integers from 1 to 8 inclusive. Judith looks at the labels of the four vertices of one of the faces of the cube. She lists these four labels in increasing order. After doing this for all six faces, she gets the following six lists: $(1,2,5,8),(3,4,6,7),(2,4,5,7),(1,3,6,8),(2,3,7,8)$, and $(1,4,5,6)$. The label of the vertex of the cube that is farthest away from the vertex labelled 2 is
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
First we choose to label one of the vertices 2 , and then label the vertex that is farthest away from this vertex $F$, as shown.

(Can you explain why the vertex labelled $F$ is the vertex that is farthest away from the vertex labelled 2?)

Each of the other six vertices of the cube lie on one of the three faces on which the vertex labelled 2 lies.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-156.jpg?height=304&width=298&top_left_y=764&top_left_x=1575)

We note that the vertex labelled $F$ is the only vertex which does not lie on a face on which the vertex labelled 2 lies.

From the six given lists, we consider those lists in which the number 2 appears.

These are: $(1,2,5,8),(2,4,5,7)$, and $(2,3,7,8)$.

Thus, the vertices labelled 1,5 and 8 lie on a face with the vertex labelled 2 , as do the vertices labelled 4,7 and 3 .

The only vertex label not included in this list is 6 .

Thus, the vertex labelled 6 is the only vertex which does not lie on a face on which the vertex labelled 2 lies.

Therefore, the correct labelling for vertex $F$, the farthest vertex from the vertex labelled 2 , is 6 .

One possible labelling of the cube is shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-156.jpg?height=304&width=355&top_left_y=1347&top_left_x=1579)

ANSWER: (D)

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