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 geometry optimization

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 $13$ visible numbers have the greatest possible sum. What is that sum?

$154$

$159$

$164$

$167$

$189$

Solution:

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 $3 \cdot 63 - 2 \cdot 12 - 1 = 189 - 24 - 1 = 164.$

Thus, C is the correct answer.

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