Difference between revisions of "2005 AMC 10A Problems/Problem 11"
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<math> \textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7 </math> | <math> \textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | Since there are <math>n^2</math> little | + | Since there are <math>n^2</math> little faces on each face of the big wooden cube, there are a total of <math>6n^2</math> little faces painted red. Moreover, as each unit cube has <math>6</math> faces, there are <math>6n^3</math> little faces in total. |
| − | + | Accordingly, as one-fourth of the little faces are painted red, we have | |
| + | <cmath>\frac{6n^2}{6n^3} = \frac{1}{4} \iff \frac{1}{n} = \frac{1}{4} \iff n = \boxed{\textbf{(B) } 4}.</cmath> | ||
| − | + | ==Solution 2== | |
| + | We recall that when a cube of side length <math>n</math> has its entire surface painted and is then split into <math>n^3</math> unit cubes, there will be exactly <math>(n-2)^3</math> unit cubes with <math>0</math> faces painted, <math>6(n-2)^2</math> unit cubes with <math>1</math> face painted, <math>12(n-2)</math> unit cubes with <math>2</math> faces painted, and <math>8</math> unit cubes with <math>3</math> faces painted. | ||
| − | <math>\ | + | (Observe that these form the terms of the binomial expansion of <math>n^3 = \left(\left(n-2\right)+2\right)^3</math>.) |
| − | < | + | Hence the total number of faces painted red is <cmath>\begin{align*}(n-2)^3 \cdot 0 + 6(n-2)^2 \cdot 1 + 12(n-2) \cdot 2 + 8 \cdot 3 &= 6\left(n^2-4n+4\right)+24n-48+24 \\ &= 6n^2-24n+24+24n-24 \\ &= 6n^2,\end{align*}</cmath> and now we may proceed as in Solution 1, again giving <math>n = \boxed{\textbf{(B) } 4}.</math> |
| − | |||
| − | <math>n=\boxed{\textbf{(B) }4}</math> | ||
==See also== | ==See also== | ||
Latest revision as of 17:31, 1 July 2025
Contents
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 1
Since there are 
little faces on each face of the big wooden cube, there are a total of 
little faces painted red. Moreover, as each unit cube has 
faces, there are 
little faces in total.
Accordingly, as one-fourth of the little faces are painted red, we have
![\[\frac{6n^2}{6n^3} = \frac{1}{4} \iff \frac{1}{n} = \frac{1}{4} \iff n = \boxed{\textbf{(B) } 4}.\]](http://latex.artofproblemsolving.com/5/3/5/535fd49981badd1829f07d9694392cb8d540869a.png)
Solution 2
We recall that when a cube of side length has its entire surface painted and is then split into unit cubes, there will be exactly 
unit cubes with 
faces painted, 
unit cubes with 
face painted, 
unit cubes with 
faces painted, and 
unit cubes with 
faces painted.
(Observe that these form the terms of the binomial expansion of 
.)
Hence the total number of faces painted red is 
and now we may proceed as in Solution 1, again giving 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
