1973 AHSME Problems/Problem 2

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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