painted cubes

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lindamat

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Apr 7, 2010
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I am stuck with the formular can anybody help me?

A cube has been painted on all faces 3x3. Then it is cut up into 27 cubelets. When counting you end up with:
12 cublets with 2 faces exposed and they use the formular 12(N-2),
6 cublets with 1 faces exposed 6(n-2)2.
I got this far by counting but I cant see where or how the formular fits in.
Thanks to all, Linda
 
Please ask someone to help you post your problem in proper English.
 
Hello, lindamat!

I think I know what you're trying to say . . .

A 3x3 cube has been painted on all faces.
Then it is cut up into 27 cubelets.
When counting you end up with:
12 cublets with 2 faces painted and they use the formula 12(N-2),
6 cublets with 1 face painted 6(n-2)2.

I got this far by counting, but I cant see where or how the formula fits in.
What formula are you referring to?
Click to expand...

\(\displaystyle \text{Consider an }n\times n\times n \text{ cube, already painted and cut up.}\)


\(\displaystyle \text{There are 8 corner cubes with three faces painted.}\)


\(\displaystyle \text{Each edge of the cube has }n-2\text{ edge-cubes with two faces painted.}\)
. . \(\displaystyle \text{The cube has twelve edges.}\)
. . \(\displaystyle \text{There are: }\:12(n-2)\text{ cubes with two faces painted.}\)


\(\displaystyle \text{Each face of the cube has }(n-2)^2\text{ face-cubes with one face painted.}\)
. . \(\displaystyle \text{The cube has six faces.}\)
. . \(\displaystyle \text{There are: }\:6(n-2)^2\text{ cubes with one face painted.}\)


\(\displaystyle \text{So exactly }where\text{ is your difficulty?}\)

 
Hi Soroban,
I have been looking and trying for over an hour to solve it and possibly just muddled up but the problem is the:
12(n - 2) and 6(n ? 2)2
what does "n" and the "2" stand for?
Thanks a lot!! Linda
 
Okay Soroban,

I want to see how do you explain - what does 2 stand for?
 
Hello again, lindamat!

Are you new to algebra?

What does "n" and the "2" stand for?
Click to expand...

\(\displaystyle \text{I said, "Consider an }n \!\times\! n \!\times\1 n\text{ cube."}\)
. . \(\displaystyle \text{So }n\text{ is the length of a side of the cube.}\)

And \(\displaystyle "2"\) stands for . . . um . . . "two".


On each face of the cube we see the following:
Code: 
    - *---*---*---*---*---*  - 
    : | C | E | E | E | C |  1
    : * - + - + - + - + - *  -
    : | E | F | F | F | E |  :
    : * - + - + - + - + - *  :
    n | E | F | F | F | E | n-2
    : * - * - * - * - * - *  :
    : | E | F | F | F | E |  :
    : * - + - + - + - + - *  -
    : | C | E | E | E | C |  1
    - *---*---*---*---*---*  -
      : 1 : -  n-2  - : 1 :

\(\displaystyle \text{There is a corner-cube (C) in each corner.}\)

\(\displaystyle \text{There are }n\!-\!2\text{ edge-cubes (E) on each edge.}\)

\(\displaystyle \text{And there are }(n\!-\!2)^2\text{ face-cubes (F) on each face.}\)


Edit: added the (F) . . .

Very funny, Denis!

Then: .\(\displaystyle n - 1 \,=\,m\;\;\hdots\;\;x + 2 \,=\,z\;\;\hdots\;\text{Ha!}\)
.
 
soroban said:
\(\displaystyle \text{There is a corner-cube (C) in each corner.}\)

\(\displaystyle \text{There are }n\!-\!2\text{ edge-cubes (E) on each edge.}\)

\(\displaystyle \text{And there are }(n\!-\!2)^2\text{ face-cubes on each face.}\)
Click to expand...
Why is there no (F) after "face-cubes"?
Why is the cube in each corner called a "corner-cube"?
Isn't n - 2 = L ? ..... L M N .....
 
Hi Soroban,
Thanks a lot for your help! In answer to your question, no I am not new to Algebra but its a LOONNGG time ago.....perhaps too long. I'll go back to my study books and start again :? . I am still lost on the N-2 but I will get there one day :lol: Cheers Linda
 
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