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

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==Solution==
 
==Solution==
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.  
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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> little faces total.  
 
Since each unit cube has <math>6</math> faces, there are <math>6n^3</math> little faces total.  

Revision as of 12:30, 30 October 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