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Difference between revisions of "1985 AHSME Problems/Problem 20"

(Created page with "==Problem== A wooden cube with edge length <math> n </math> units (where <math> n </math> is an integer <math> >2 </math>) is painted black all over. By slices parallel to it...")
 
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==Problem==
 
==Problem==
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A wooden [[cube]] with edge length <math> n </math> units (where <math> n </math> is an integer <math> >2 </math>) is painted black all over. By slices parallel to its faces, the cube is cut into <math> n^3 </math> smaller cubes each of unit length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is <math> n </math>?
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A wooden cube with edge length <math>n</math> units (where <math>n</math> is an integer <math>>2</math>) is painted black all over. By slices parallel to its faces, the cube is cut into <math>n^3</math> smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is <math>n</math>?
  
 
<math> \mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \  } 7 \qquad \mathrm{(D) \  } 8 \qquad \mathrm{(E) \  }\text{none of these} </math>
 
<math> \mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \  } 7 \qquad \mathrm{(D) \  } 8 \qquad \mathrm{(E) \  }\text{none of these} </math>
  
 
==Solution==
 
==Solution==
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Notice that if we remove the outer layer of unit cubes from the entire cube, we're left with a smaller cube of side length <math> n-2 </math>. Notice also that this contains all of the unpainted cubes and nothing else, so there are <math> (n-2)^3 </math> unpainted cubes. Also notice that if we take one face of the cube and remove the outer edge, we're left with a square of side length <math> n-2 </math> containing all of the cubes on that face with exactly one face painted. This can be done for the other <math> 5 </math> faces as well, for a total of <math> 6(n-2)^2 </math> cubes with only one face painted.  
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Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length <math>(n-2)</math>, which contains all of the unpainted cubes and no others. This shows that there are exactly <math>(n-2)^3</math> unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length <math>(n-2)</math> containing all of the cubes on that face with exactly one face painted. Making the same argument for the other <math>5</math> faces as well, we deduce that there are a total of <math>6(n-2)^2</math> cubes with only one face painted.  
  
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<math> 6(n-2)^2=(n-2)^3 </math>
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Accordingly, we require <cmath>\begin{align*}(n-2)^3 = 6(n-2)^2 &\iff n-2 = 6 \qquad \text{(as } n > 2\text{, so } n-2 \neq 0\text{)} \\ &\iff n = \boxed{\text{(D)} \ 8}.\end{align*}</cmath>
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<math> 6=n-2 </math>
 
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<math> n=8, \boxed{\text{D}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=19|num-a=21}}
 
{{AHSME box|year=1985|num-b=19|num-a=21}}
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{{MAA Notice}}

Latest revision as of 22:17, 19 March 2024

Problem

A wooden cube with edge length $n$ units (where $n$ is an integer $>2$) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit edge length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $n$?

$\mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ }\text{none of these}$

Solution

Observe that if we remove the outer layer of unit cubes from the entire cube, what remains is a smaller cube of side length $(n-2)$, which contains all of the unpainted cubes and no others. This shows that there are exactly $(n-2)^3$ unpainted cubes. Similarly, taking one face of the cube and removing the outer edge leaves a square of side length $(n-2)$ containing all of the cubes on that face with exactly one face painted. Making the same argument for the other $5$ faces as well, we deduce that there are a total of $6(n-2)^2$ cubes with only one face painted.

Accordingly, we require \begin{align*}(n-2)^3 = 6(n-2)^2 &\iff n-2 = 6 \qquad \text{(as } n > 2\text{, so } n-2 \neq 0\text{)} \\ &\iff n = \boxed{\text{(D)} \ 8}.\end{align*}

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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