Question

A cube is painted (all 6 faces) and cut into one-inch cubes. If the original cube has n inch edges, how many (in terms of n) one-inch cubes have:

a. exactly 3 faces painted?
b. exactly 2 faces painted?
c. exactly 1 face painted?
d. exactly 0 faces painted?

e. What is the total number of one-inch cubes?

Answer 1

a. 8

b. 12(n -2)

c. 6(n -2)^2

d. (n -2)^3

Explanation:

A cube has 8 vertices (corners), 12 edges, and 6 faces.

a. Three faces will be painted on each of the 8 corner cubes.

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b. Two faces will be painted on the edge cubes that are not corner cubes. For an n-inch edge, n-2 inches of it are not part of the corner cubes. Two faces will be painted on 12(n -2) edge cubes.

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c. One face will be painted on each face cube that is not part of an edge or a corner. For a face of dimensions n inches square, the cubes of interest comprise a square that is n-2 inches on a side. There are 6 such faces of the larger cube. One face will be painted on 6(n -2)^2 face cubes.

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d. The cubes internal to the larger cube that are not part of any face, edge, or corner make up a cube that is (n -2)^3 smaller cubes. This is the number with no paint.

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e. Of course the total number of cubes with or without paint is n^3. This is the sum of the cubes in each category:

(((n -2) +6)(n -2) +12)(n -2) +8 = ((n +4)(n -2) +12)(n -2) +8

= (n^2 +2n +4)(n -2) +8

= n^3 +2n^2 -2n^2 +4n -4n -8 +8

= n^3