[Guest]

Take this 10 x 10 x 10 cube and coat it in paint. (so that only the outside faces get painted)

Now, disassemble the cube and count how many of the blocks have...

NO Faces painted: ?

ONE Face painted: ?

TWO Faces painted: ?

THREE Face painted: ?

Now determine a formula to count the number of faces with each number of sides painted as stated above.

- It must work for ANY cube formed the same way as this one (of smaller blocks) that is at least 3 x 3 x 3

Just starting you out with a simple one, I may post more with more complicated shapes than just a cube (depending on if I can actually draw them so they look proper )

(please don't over-analyze the paint it is just meant to mark the faces, don't say something like: Well the paint wasn't dry so you have to subtract 10 painted faces for each of your fingers because you wiped off all the paint on certain blocks)

- K4D

[Guest]

3 faces - always just the corners so 8 regardless of size.

1 face - 6[HxW-4(H-1)]

2 face - 12(H-2)

Close?

edit: No Faces - Just HxWxL - (#1face + #2face + #3face)

Edited April 28, 2009 by palmerc7

[Guest]

No faces - 512

1 face - 384

2 faces - 96

3 faces - 8

[Guest]

I'm just guessing because I did this in my head really quickly but...

8 with 3 sides painted

96 with 2 sides painted

384 with 1 side painted

and 512 with no sides painted

I don't know if their right, but they sound good enough to me.

[Guest]

8 blocks with 3 painted faces on any size cube

(length of side minus 2)*12 number of blocks with 2 sides painted

[(length of side minus 2)^2]*6 for the number of blocks with 1 face painted

(length of side^3)-total number of partially painted blocks to find how many have no paint

[Guest]

Yes i think idid it

there r 400 with 0 sides painted

360 with one side painted

216 with 2 sides painted

24 with 3 sides painted

[Guest]

My first post...

Let n = the number of units of one side. In the example: n = 10.

0 faces= 512 (n-2)^3

1 face= 384 (n-2)^2 * 6

2 faces= 96 (n-2) *12

3 faces= 8 always 8

[Joe's Student]

3 faces - always just the corners so 8 regardless of size.

1 face - 6[HxW-4(H-1)]

2 face - 12(H-2)

Close?

edit: No Faces - Just HxWxL - (#1face + #2face + #3face)

Alternatively:

0 Faces = (H-2)x(W-2)x(L-2)

[Guest]

Well I've seen the answer posted several times already so i'll go ahead and post what I got.

In a cube consisting of n^3 blocks the formulas are:

NO Faces: (n-2)^3

ONE Face: 6(n-2)^2

TWO Faces: 12(n-2)

THREE Faces: 8

So these formulas yield:

512 blocks with NO faces painted.

384 blocks with ONE face painted.

96 blocks with TWO faces painted.

8 blocks with THREE faces painted.

You can check this by summing your results:

512 + 384 + 96 + 8 = 1000 = 10^3

[Joe's Student]

A puzzle like this but with more complex shapes would be interesting .

[Guest]

A puzzle like this but with more complex shapes would be interesting .

Thats exactly what I was about to start working on, but as I think about the shapes I don't see many that wouldn't either be similar to the cube, or require me to reveal at least strong hints to the formulas while revealing how the shape was constructed