2008 AMC 12A Problem 11

Below is the professionally curated solution for Problem 11 of the 2008 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12A solutions, or check the answer key.

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Concepts:cube geometryoptimization

Difficulty rating: 1560

11.

Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the 1313 visible numbers have the greatest possible sum. What is that sum?

154154

159159

164164

167167

189189

Solution:

The six faces of each cube sum to 1+2+4+8+16+32=63.1 + 2 + 4 + 8 + 16 + 32 = 63. From the pattern, the pairs of opposite faces are 11 & 32,32, 22 & 16,16, and 44 & 8.8.

Each of the two lower cubes hides a pair of opposite faces (top and bottom); hiding the pair 4+8=124 + 8 = 12 is best. The top cube hides only its bottom face, so hide the 1.1.

The greatest sum is 3632121=189241=164. 3 \cdot 63 - 2 \cdot 12 - 1 = 189 - 24 - 1 = 164.

Thus, C is the correct answer.

Problem 11 in Other Years