2007 AMC 10A Problem 11

Question

The numbers from

1

to

8

are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

14

16

18

20

24

Each vertex of a cube lies on exactly three faces

Adding all six face sums counts each vertex

3

times:

3(1 + 2 + \cdots + 8)

Each vertex belongs to exactly three faces, so summing the numbers over all six faces gives

\begin{aligned} 3(1 + 2 + \cdots + 8) &= 3 \cdot 36 \\ &= 108. \end{aligned}

There are six faces, so the common sum is

108 \div 6 = 18.

Thus, the correct answer is C.

Problem 11 in other years