2008 AMC 12A Problems/Problem 11

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?

$[asy] unitsize(.8cm); pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label("2",(-0.5,0.5)); draw(unitsquare,p); label("32",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label("16",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label("4",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label("8",(0.5,-0.5)); [/asy]$

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

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

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