Difference between revisions of "1987 AJHSME Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
− | Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus <math>27 - 7 = | + | Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus <math>27 - 7 = 20</math> our answer is <math>\boxed{\text{C}}</math>. |
~Shadow-18 | ~Shadow-18 |
Revision as of 19:04, 30 April 2021
Contents
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 unit cube has more than one face painted, so the number of unit 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 .
Solution 2
Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus our answer is .
~Shadow-18
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 |
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