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 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 .
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.