The entire exterior of a large wooden cube is painted red

problem never specifies a real number for any of the steps, so select a smart number for n, say n = 3. In this case, the original cube is painted and then cut into a 3 × 3 × 3 assemblage of 27 smaller cubes. Imagine the top face of the original (large) cube (if you know what a Rubik’s Cube is, picture one!). Every cube on that face has been painted on at least one side. The same is true for the bottom face. Now think about the middle “slice” of the cube. This “slice” contains a total of 9 cubes, but only the one in the very middle has not been painted on any of its sides.



Therefore, only one of those 27 cubes—the one in the middle of the structure—remains unpainted. When n = 3, the answer is 26.

Plug n = 3 into the answers and look for 26:

(A) 6(32) = 6(9) = 54
(B) 6(32) – 12(3) + 8 = 54 – 36 + 8 = 26
(C) 6(32) – 16(3) + 24 = 54 – 48 + 24 = 30
(D) 4(32) = 4(9) = 36
(E) 24(3) – 24 = 72 – 24 = 48

How is 6n^2 =54 if n=3?

Thanks for your help!

6n^2 = 6*3^2 = 6*9 = 54.
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if we have a Cube and it is cut into (N)^3 smaller cubes -------> then the Side Length must be N

ex: if we cut a Cube into (3)^3 = 27 cubes of Equal Length

we could each have the 27 Cubes have the following dimensions ----> 1 x 1 x 1

putting the 27 cubes together to create 1 Larger Cube that we originally cut from -----> 3 x 3 x 3

Side Length = 3 = N ------ the Cube Root of (N)^3


you could use the same logic for the example of taking a Larger Cube and cutting it into (N)^3 = (4)^3 = 64 Cubes

again, each smaller cube's dimensions ------> 1 x 1 x 1

putting the 64 smaller cubes of dimension 1 x 1 x 1 together to create the Original Cube we cut from -----> you would get the Original Un-Cut Cube with Edge Length = 4 = N



Given the Larger Cube has Dimensions N x N x N -------> and we cut the Cube into (N)^3 Number of Cubes

No. of Cubes that have 3 Faces Painted ----> on the 8 Vertices = 8

No. of Cubes that have 2 Faces Painted ------> (12 Edges) * (N Cubes on each Edge - 2 Cubes on Each Vertex) = 12(N - 2)

No. of Cubes that have 1 Face Painted -------> (6 Faces) * (N Cubes along 1 dimension of a Face - 2 Cubes) * (N Cubes along Other Dimension of a particular Face - 2 Cubes) = 6 * (N - 2) * (N - 2) = 6 * (N - 2)^2


No. of Cubes that have AT LEAST 1 Face Painted =

8 + (12)(N-2) + (6)(N - 2)^2 =

8 + 12N - 24 + 6(N)^2 - 24N + 24 =

6(N)^2 - 12N + 8

-B-

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