Difference between revisions of "2008 AMC 12A Problems/Problem 11"
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So, the minimum sum of the <math>5</math> unexposed faces is <math>2\cdot12+1=25</math>. Since the sum of the numbers on all the cubes is <math>3(32+16+8+4+2+1)=189</math>, the maximum possible sum of <math>13</math> visible numbers is <math>189-25=164 \Rightarrow C</math>. | So, the minimum sum of the <math>5</math> unexposed faces is <math>2\cdot12+1=25</math>. Since the sum of the numbers on all the cubes is <math>3(32+16+8+4+2+1)=189</math>, the maximum possible sum of <math>13</math> visible numbers is <math>189-25=164 \Rightarrow C</math>. | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ~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== | ==See Also== | ||
{{AMC12 box|year=2008|num-b=10|num-a=12|ab=A}} | {{AMC12 box|year=2008|num-b=10|num-a=12|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
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 visible numbers have the greatest possible sum. What is that sum?
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. Two cubes have 4 exposed faces. 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 top cube has one unexposed face, so use 1 as the unexposed face.
~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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.