2008 AMC 12A Problems/Problem 11
Contents
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 
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]](http://latex.artofproblemsolving.com/a/3/2/a32f7c9839807c1cb204fd5a2f87f33747244a18.png)
Solution
To maximize the sum of the faces that are showing, we can minimize the sum of the numbers of the 
faces that are not showing.
The bottom 
cubes each have a pair of opposite faces that are covered up. When the cube is folded, 
; 
; and 
are opposite pairs. Clearly 
has the smallest sum.
The top cube has 1 number that is not showing. The smallest number on a face is 
.
So, the minimum sum of the unexposed faces is 
. Since the sum of the numbers on all the cubes is 
, the maximum possible sum of visible numbers is 
.
Solution 2
Conversely, maximize the sum. The first two cubes have 4 numbers out facing. Since 
, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but 
is the greatest at 51. There are 2 such cubes, so 51*2. The cube on top only hides 1 number, so simply use 1 as the unexposed face. 
~BJHHar
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| 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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