Difference between revisions of "2008 AMC 12A Problems/Problem 11"

Latest revision as of 15:50, 24 September 2024

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?

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

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

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 $189-25=164 \Rightarrow C$ .

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. $2(51)+32+16+8+4+2=164 \implies \boxed{\textbf{(C)} 164}$

~BJHHar

Solution 3

Two cubes have 4 exposed faces in the same configuration (a ring of 4 squares around the cube). Since the numbers are the same on each cube, they will have the same maximal sum, which is at least 32+16 = 48. The third cube has 5 visible faces. Obviously, the non-visible face on this cube will be the smallest, which is 1. So, this third cube contributes 32+16+8+4+2 = 62. The sum is then at least 62 + 48 + 48. The sum must be even because 62 is even and there are two cubes with the same sum. The only answer choice left is $\boxed{\textbf{(C)} 164}$

@@ Line 32: / Line 32: @@
 ==Solution 2==
-Conversely, maximize the sum. Two cubes have 4 exposed faces. Since <math>32>16+8+4+2</math>, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but <math>(32,16,2,1)</math> 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. <math>2(51)+32+16+8+4+2=164 \implies \boxed{\textbf{(C)}164}</math>
+Conversely, maximize the sum. Two cubes have 4 exposed faces. Since <math>32>16+8+4+2</math>, 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but <math>(32,16,2,1)</math> 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. <math>2(51)+32+16+8+4+2=164 \implies \boxed{\textbf{(C)} 164}</math>
 ~BJHHar
+==Solution 3==
+Two cubes have 4 exposed faces in the same configuration (a ring of 4 squares around the cube). Since the numbers are the same on each cube, they will have the same maximal sum, which is at least 32+16 = 48.
+The third cube has 5 visible faces. Obviously, the non-visible face on this cube will be the smallest, which is 1. So, this third cube contributes 32+16+8+4+2 = 62.
+The sum is then at least 62 + 48 + 48. The sum must be even because 62 is even and there are two cubes with the same sum. The only answer choice left is <math>\boxed{\textbf{(C)} 164}</math>
 ==See Also==
 {{AMC12 box|year=2008|num-b=10|num-a=12|ab=A}}
 {{MAA Notice}}

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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