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Difference between revisions of "1998 AJHSME Problems/Problem 21"

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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==
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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.
Line 16: Line 16:
  
 
<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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 +
You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
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 +
There are <math>4</math> 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, <math>4</math> more cubes are removed.
 +
 +
Finally, you repeat this on the third layer, for <math>4</math> 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 <math>4 + 4 + 4 = 12</math> cubes of the <math>4 \times 4 \times 4 = 64</math> cubes.
 +
 +
Thus, <math>64 - 12 = 52</math> cubes remain, and the answer is <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 11:15, 31 July 2011

Problem 21

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?

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

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 $\boxed{B}$.

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
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