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> | ||
| − | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== | ||
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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>. | ||
| + | |||
| + | ===Solution 3=== | ||
| + | We can use casework to solve this. | ||
| + | |||
| + | Case <math>1</math>: The cubes on the sides | ||
| + | |||
| + | 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). | ||
| + | So there are <math>16 \times 4-16=64-16=48</math> cubes on the sides. | ||
| + | |||
| + | Case <math>2</math>: The cubes on the bottom | ||
| + | |||
| + | 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. | ||
| + | |||
| + | <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]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:55, 8 February 2023
Problem
A 
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?
Solutions
Solution 1
Each small cube would have dimensions 
making each cube a unit cube.
If there are 
cubes per face and there are 
faces we are counting, we have 
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.
Solution 2
You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
There are 
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, more cubes are removed.
Finally, you repeat this on the third layer, for 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 
cubes of the 
cubes.
Thus, 
cubes remain, and the answer is 
.
Solution 3
We can use casework to solve this.
Case 
: The cubes on the sides
There are four sides of a cube which means 
cubes, but there will be an overlap of cubes ( sides 
cubes on each side).
So there are 
cubes on the sides.
Case 
: The cubes on the bottom
There are cubes on the bottom, but by counting the cubes on the sides, 
of the cubes are already accounted for. 
cubes on the bottom that aren't already counted.
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. 
