2007 AMC 10A Problem 11

Problem 11 of 25IntermediateGeometry Combinatorics

The numbers from

1

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?

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Solution

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.

Tagged: double counting · cube geometry