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Difference between revisions of "1973 AHSME Problems/Problem 2"

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===Solution 1===
 
===Solution 1===
  
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The total number of cubes is S10^3 or 1000S. Because each surface of the large cube is one cube deep, the number of the unpainted cubes is <math>8^3 = 512</math>, since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are <math>1000-512=\boxed{\textbf{(C) } 488}</math> cubes that have at least one face painted.
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The total number of cubes is <math>10^3</math> or <math>1000</math>. Because each surface of the large cube is one cube deep, the number of the unpainted cubes is <math>8^3 = 512</math>, since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are <math>1000-512=\boxed{\textbf{(C) } 488}</math> cubes that have at least one face painted.
  
 
===Solution 2===
 
===Solution 2===
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==See Also==
 
==See Also==
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{{AHSME 35p box|year=1973|num-b=1|num-a=3}}
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{{AHSME 30p box|year=1973|num-b=1|num-a=3}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Latest revision as of 12:56, 20 February 2020

Problem

One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is

$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$

Solutions

Solution 1

The total number of cubes is $10^3$ or $1000$. Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$, since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\boxed{\textbf{(C) } 488}$ cubes that have at least one face painted.

Solution 2

Each face has $100$ cubes, so multiply by six to get $600$. However, we overcounted each small cube on the edge but not on corner of the big cube once and each small cube on the corner of the big cube twice. Thus, there are $600 - (12 \cdot 8 + 2 \cdot 8) = \boxed{\textbf{(C) } 488}$ cubes that have at least one face painted.

See Also

1973 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AHSME Problems and Solutions
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