Difference between revisions of "1987 AJHSME Problems/Problem 7"
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Clearly no cube has more than one face painted. Therefore, the number of cubes with at least one face painted is equal to the number of painted unit squares. | Clearly no cube has more than one face painted. Therefore, the number of cubes with at least one face painted is equal to the number of painted unit squares. | ||
− | There are <math>10</math> painted unit squares on the half of the cube shown, so there are <math>10\cdot 2=20</math> cubes with at least one face painted | + | There are <math>10</math> painted unit squares on the half of the cube shown, so there are <math>10\cdot 2=20</math> unit cubes with at least one face painted, thus our answer is <math>\boxed{\text{C}}</math>. |
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− | <math>\boxed{\text{C}}</math> | ||
==See Also== | ==See Also== |
Revision as of 13:51, 28 October 2020
Problem
The large cube shown is made up of identical sized smaller cubes. For each face of the large cube, the opposite face is shaded the same way. The total number of smaller cubes that must have at least one face shaded is
Solution
Clearly no cube has more than one face painted. Therefore, the number of cubes with at least one face painted is equal to the number of painted unit squares.
There are painted unit squares on the half of the cube shown, so there are unit cubes with at least one face painted, thus our answer is .
See Also
1987 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.