2008 AMC 12A Problems/Problem 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?

[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]

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


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 $189-25=164 \Rightarrow C$.

Solution 2

Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed face. $2(51)+32+16+8+4+2=164 \implies \boxed{\textbf{(C)} 164}$


See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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