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Difference between revisions of "1998 AJHSME Problems/Problem 21"
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==Problem 21==
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==Problem==
  
 
A <math>4\times 4\times 4</math> cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
 
A <math>4\times 4\times 4</math> cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
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<math>\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80</math>
 
<math>\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80</math>
  
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==Solution 1==
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==Solutions==
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===Solution 1===
  
 
Each small cube would have dimensions <math>1\times 1\times 1</math> making each cube a unit cube.
 
Each small cube would have dimensions <math>1\times 1\times 1</math> making each cube a unit cube.
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<math>80-(4\times 4+12)=80-28=52=\boxed{B}</math>
 
<math>80-(4\times 4+12)=80-28=52=\boxed{B}</math>
  
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==Solution 2==
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===Solution 2===
  
 
You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
 
You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
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Thus, <math>64 - 12 = 52</math> cubes remain, and the answer is <math>\boxed{B}</math>.
 
Thus, <math>64 - 12 = 52</math> cubes remain, and the answer is <math>\boxed{B}</math>.
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===Solution 3===
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We can use casework to solve this.
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Case <math>1</math>: The cubes on the sides
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There are four sides of a cube which means <math>64</math> cubes, but there will be an overlap of <math>16</math> cubes (<math>4</math> sides <math>\times</math> <math>4</math> cubes on each side).
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So there are <math>16 \times 4-16=64-16=48</math> cubes on the sides.
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Case <math>2</math>: The cubes on the bottom
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There are <math>16</math> cubes on the bottom, but by counting the cubes on the sides, <math>12</math> of the cubes are already accounted for. <math>16-12=4</math> cubes on the bottom that aren't already counted.
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<math>48+4=\boxed{\textbf{(B)} 52}</math> cubes touching at least one side. Since there is no top, we don't need to count the top face cubes.
  
 
== See also ==
 
== See also ==
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* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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{{MAA Notice}}

Latest revision as of 17:55, 8 February 2023

Problem

A $4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box. How many of these small cubes touch a side or the bottom of the box?

$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solutions

Solution 1

Each small cube would have dimensions $1\times 1\times 1$ making each cube a unit cube.

If there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\times 5= 80$ cubes.

Some cubes are on account of overlap between different faces.

We could reduce this number by subtracting the overlap areas, which could mean subtracting 4 cubes from each side and 12 from the bottom.

$80-(4\times 4+12)=80-28=52=\boxed{B}$

Solution 2

You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".

There are $4$ cubes in the center of the top face that do not fit the description. Remove those.

Once you remove those cubes on top, you must go down one level and remove the cubes in the same position on the second layer. Thus, $4$ more cubes are removed.

Finally, you repeat this on the third layer, for $4$ more cubes.

Once you do the top three layers, you will be on the bottom layer, and you don't remove any more cubes.

This means you removed $4 + 4 + 4 = 12$ cubes of the $4 \times 4 \times 4 = 64$ cubes.

Thus, $64 - 12 = 52$ cubes remain, and the answer is $\boxed{B}$.

Solution 3

We can use casework to solve this.

Case $1$: The cubes on the sides

There are four sides of a cube which means $64$ cubes, but there will be an overlap of $16$ cubes ($4$ sides $\times$ $4$ cubes on each side). So there are $16 \times 4-16=64-16=48$ cubes on the sides.

Case $2$: The cubes on the bottom

There are $16$ cubes on the bottom, but by counting the cubes on the sides, $12$ of the cubes are already accounted for. $16-12=4$ cubes on the bottom that aren't already counted.

$48+4=\boxed{\textbf{(B)} 52}$ cubes touching at least one side. Since there is no top, we don't need to count the top face cubes.

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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