Question

Each vertex of a cube is to be labeled with an integer from $1$ through $8,$ with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

$1$

$3$

$6$

$12$

$24$

Solution:

Each vertex belongs to $3$ faces, so $6S=3(1+2+\cdots+8)=108,$ giving each face-sum $S=18.$

The four-element subsets containing $1$ with sum $18$ are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},$ and $\{1,4,6,7\}.$ Three of these contain both $1$ and $8,$ so $1$ and $8$ must lie on two adjacent vertices.

Rotate the cube so that $1$ is at the lower-left-front vertex and $8$ at the lower-right-front vertex. The numbers $4,6,7$ must label the remaining vertices of the face containing $1,$ which can be done in $3!=6$ ways; then $5,3,2$ are forced onto the opposite vertices. Hence there are $6$ arrangements.

Thus, the correct answer is C.

2016 AMC 12A Problem 14

Solution:

Problem 14 in Other Years