Painted Cube Faces

Probability Level 3

Consider an $n \times n \times n$ cube $(n \geq 3)$ whose outer surface is painted blue. The unit cubes at the corners each have 3 blue faces, while the unit cubes in the center each have 0 blue faces.

For what value of $n$ will the number of unit cubes with 0 blue faces be twice the number of unit cubes with 1 blue face?

The answer is 14.

1 solution

Chung Kevin
Nov 11, 2016

For $n \geq 3$ ,
The number of cubes with only one blue face, are those that lie on the face of the large cube, but not the edges. There are 6 faces, and each of them have $(n-2) \times (n-2)$ such cubes.

The number of cubes with no blue faces, are those that lie completely within the large cube. There are $(n-2) \times (n-2) \times (n-2)$ of them.

Hence, we have to solve $2 \times 6 \times (n-2)^2 = (n-2)^3$ .
This gives us $(n-2) ^2 \times \left( 12 - (n-2) \right) = 0$ , so $n = 2, 14$ .

Since we have $n \geq 3$ (to remove the trivial cases), thus the answer is $n = 14$ .

According to the question, we are trying to solve 6 (n-2)^2 = 2 (n-3)^3, not the otherway round. The number of cubes with no blue face is supposed to be double the numer of cubes with one blue face. You are calculating the opposite of the question.

tim waring - 4 years, 7 months ago

Are you sure? If the number of cubes with no blue face is $1728$ , then we want the number of cubes with 1 blue face to be $864$ .

Chung Kevin - 4 years, 7 months ago

Painted Cube Faces

The answer is 14.

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