Difference between revisions of "2005 AMC 10A Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | Since there are <math>n^2</math> | + | Since there are <math>n^2</math> little [[face]]s on each face of the big wooden [[cube (geometry) | cube]], there are <math>6n^2</math> little faces painted red. |
− | Since each unit cube has <math>6</math> faces, there are <math>6n^3</math> faces. | + | Since each unit cube has <math>6</math> faces, there are <math>6n^3</math> little faces total. |
− | Since one-fourth of the faces are painted red, | + | Since one-fourth of the little faces are painted red, |
<math>\frac{6n^2}{6n^3}=\frac{1}{4}</math> | <math>\frac{6n^2}{6n^3}=\frac{1}{4}</math> | ||
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<math>\frac{1}{n}=\frac{1}{4}</math> | <math>\frac{1}{n}=\frac{1}{4}</math> | ||
− | <math>n=4\ | + | <math>n=4\Longrightarrow \mathrm{(B)}</math> |
− | ==See | + | ==See also== |
− | + | {{AMC10 box|year=2005|ab=A|num-b=10|num-a=12}} | |
− | + | {{MAA Notice}} | |
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Revision as of 16:36, 8 December 2020
Problem
A wooden cube units on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?
Solution
Since there are little faces on each face of the big wooden cube, there are little faces painted red.
Since each unit cube has faces, there are little faces total.
Since one-fourth of the little faces are painted red,
See also
2005 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.