2008 AMC 12A Problems/Problem 11

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Problem

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?

$\mathrm{(A)}\ 154\qquad\mathrm{(B)}\ 159\qquad\mathrm{(C)}\ 164\qquad\mathrm{(D)}\ 167\qquad\mathrm{(E)}\ 189$

Solution

To maximize the sum of the $13$ faces that are showing, we can minimize the sum of the numbers of the $5$ faces that are not showing.

The bottom $2$ cubes each have a pair of opposite faces that are covered up. When the cube is folded, $(1,32)$ ; $(2,16)$ ; and $(4,8)$ are opposite pairs. Clearly $4+8=12$ has the smallest sum.

The top cube has 1 number that is not showing. The smallest number on a face is $1$ .

So, the minimum sum of the $5$ unexposed faces is $2\cdot12+1=25$ . Since the sum of the numbers on all the cubes is $3(32+16+8+4+2+1)=189$ , the maximum possible sum of $13$ visible numbers is $189-25=164 \Rightarrow C$ .

2008 AMC 12A (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
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2008 AMC 12A Problems/Problem 11

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Problem

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