Difference between revisions of "2005 AMC 10A Problems/Problem 11"

(added problem and solution)
 
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==Solution==
 
==Solution==
Since there are <math>n^2</math> faces on each face of the wooden cube, there are <math>6n^2</math> faces painted red.  
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Since there are <math>n^2</math> little [[face]]s on each face of the big wooden [[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.  
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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,  
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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\Rightarrow B</math>
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<math>n=4\Longrightarrow \mathrm{(B)}</math>
  
 
==See Also==
 
==See Also==
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*[[2005 AMC 10A Problems/Problem 12|Next Problem]]
 
*[[2005 AMC 10A Problems/Problem 12|Next Problem]]
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[[Category:Introductory Geometry Problems]]

Revision as of 10:54, 2 August 2006

Problem

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?

$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7$

Solution

Since there are $n^2$ little faces on each face of the big wooden cube, there are $6n^2$ little faces painted red.

Since each unit cube has $6$ faces, there are $6n^3$ little faces total.

Since one-fourth of the little faces are painted red,

$\frac{6n^2}{6n^3}=\frac{1}{4}$

$\frac{1}{n}=\frac{1}{4}$

$n=4\Longrightarrow \mathrm{(B)}$

See Also

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