Well I'm stuck on this one math problem in my homework . I tried looking on hotmath, but sadly its a classwork problem and I have no clue how to do this or where to start So hopefully you guys can help me out
-Thanks!!!
Heres the problem:
Well I'm stuck on this one math problem in my homework . I tried looking on hotmath, but sadly its a classwork problem and I have no clue how to do this or where to start So hopefully you guys can help me out
-Thanks!!!
Heres the problem:
Hello, scorpio!
I'll reword the problem slightly ... for convenience.
An $\displaystyle n \times n \times n$ cube is painted, then it is cut into $\displaystyle n^3$ unit cubes.
A unit cube can have: 0, 1, 2, or 3 painted faces.
Making some sketches, the table can be extended and generalized.
. . . . . . . . $\displaystyle \text{No. of painted faces}$
. . $\displaystyle \begin{array}{c||c|c|c|c|} \hline
n & \quad3\quad & 2 & 1 & 0 \\ \hline \hline
2 & 8 & 0 & 0 & 0 \\
3 & 8 & 12 & 6 & 1 \\
4 & 8 & 24 & 24 & 8 \\
5 & 8 & 36 & 54 & 27 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
n & 8 & 12(n-2) & 6(n-2)^2 & (n-2)^3 \\ \hline
\end{array}$
Explanation
We have an $\displaystyle n\times n\times n$ cube.
There are always 8 corner-cubes (3 painted faces).
There are 12 edges; each edge has $\displaystyle (n-2)$ edge-cubes (2 painted faces).
There are 6 faces; each face has $\displaystyle (n-2)^2$ face-cubes (1 painted face).
Inside, there is a smaller cube of side $\displaystyle (n-2)\!:\;\;(n-2)^3$ unpainted cubes.