Question

Each vertex of a cube is to be labeled with an integer $1$ 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:

Opposite faces together use all eight labels, whose sum is $36$ , so every face must have sum $18$ .

Put label $1$ at one vertex, and let the three adjacent labels be $a,b,c$ . The three vertices adjacent to those across faces are then forced to be $17-a-b,\quad 17-a-c,\quad 17-b-c,$ and the opposite vertex is $a+b+c-16$ .

Assume the three neighbor labels are listed in increasing order. Checking possible triples from $2,3,\ldots,8$ , the valid triples are $(4,6,8)$ , $(4,7,8)$ , and $(6,7,8)$ . Each gives two non-rotationally equivalent arrangements, so there are $3\cdot2=6$ arrangements.

Thus, the correct answer is C.

2016 AMC 10A Problem 18

Solution:

Problem 18 in Other Years