Difference between revisions of "2005 AMC 12A Problems/Problem 10"
(New page: == Problem == A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of...) |
(→See also) |
||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>? | + | A wooden [[cube]] <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>? |
<math> | <math> | ||
Line 6: | Line 6: | ||
</math> | </math> | ||
== Solution == | == Solution == | ||
− | There are <math>6n^3</math> sides total, and <math>6n^2</math> are painted red. | + | There are <math>6n^3</math> sides total on the unit cubes, and <math>6n^2</math> are painted red. |
<math>\dfrac{6n^2}{6n^3}=\dfrac{1}{4} \Rightarrow n=4 \rightarrow \mathrm {B}</math> | <math>\dfrac{6n^2}{6n^3}=\dfrac{1}{4} \Rightarrow n=4 \rightarrow \mathrm {B}</math> | ||
− | |||
− | |||
== See also == | == See also == | ||
Line 16: | Line 14: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 20:19, 3 July 2013
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
There are sides total on the unit cubes, and are painted red.
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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.