The entire exterior of a large wooden cube is painted red, and then the cube is sliced into \(n^3\) smaller cubes (where

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The entire exterior of a large wooden cube is painted red, and then the cube is sliced into \(n^3\) smaller cubes (where \(n > 2\)). Each of the smaller cubes is identical. In terms of \(n,\) how many of these smaller cubes have been painted red on at least one of their faces?

A. \(6n^2\)
B. \(6n^2 - 12n + 8\)
C. \(6n^2 - 16n + 24\)
D. \(4n^2\)
E. \(24n - 24\)

Answer: B

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Gmat_mission wrote:
Tue Nov 10, 2020 8:14 am
The entire exterior of a large wooden cube is painted red, and then the cube is sliced into \(n^3\) smaller cubes (where \(n > 2\)). Each of the smaller cubes is identical. In terms of \(n,\) how many of these smaller cubes have been painted red on at least one of their faces?

A. \(6n^2\)
B. \(6n^2 - 12n + 8\)
C. \(6n^2 - 16n + 24\)
D. \(4n^2\)
E. \(24n - 24\)

Answer: B

Solution:

The number of smaller cubes that have no faces painted is (n - 2)^3. Therefore, the number of smaller cubes that have at least one face painted is:

n^3 - (n - 2)^3 = n^3 - (n^3 - 6n^2 + 12n - 8) = 6n^2 - 12n + 8

Alternate Solution:

Let n = 3. Then we know we have 3^3 = 27 smaller cubes, and 26 of them (all except the innermost cube) will have at least one face that is painted red.

If we plug in 3 for n in each answer choice, we see that choice B is the only one that gives 26 as the answer: 6 * 9 - 12 * 3 + 8 = 54 - 36 + 8 = 26.

Answer: B


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