2008 AMC 12A Problem 11

Question

Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the

13

visible numbers have the greatest possible sum. What is that sum?

154

159

164

167

189

On each cube the opposite faces are

1

&

32,

2

&

16,

and

4

&

8

The two lower cubes each hide a pair of opposite faces; the top cube hides only its bottom face

The six faces of each cube sum to

1 + 2 + 4 + 8 + 16 + 32 = 63.

From the pattern, the pairs of opposite faces are

1

&

32,

2

&

16,

and

4

&

8.

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair

4 + 8 = 12

is best. The top cube hides only its bottom face, so hide the

1.

The greatest sum is

\begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned}

Thus, C is the correct answer.

Problem 11 in other years