2008 AMC 12A Problem 11

Attempt Problem 11 of the 2008 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Answer: C
Concepts:cube geometryoptimization
Difficulty rating: 1560
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. \begin{aligned} &3 \cdot 63 - 2 \cdot 12 - 1 \\ &= 189 - 24 - 1 \\ &= 164. \end{aligned}

Thus, C is the correct answer.

← Problem 10#10
Full Exam

Problem 11 in Other Years