• January 8th 2011, 05:27 AM
abm
For a 10x10x10 cube, 1000 cubes total (10^3), and 8^3 don't have paint on them.
And what fraction of the cubes would have a painted side upward?
• January 8th 2011, 06:33 AM
Quote:

Originally Posted by abm
For a 10x10x10 cube, 1000 cubes total (10^3), and 8^3 don't have paint on them.
And what fraction of the cubes would have a painted side upward?

I guess we assume the $8^3$ cubes are the "internal ones",
since that is the number of cubes whose faces are not visible
(think of an 8x8x8 cube covered by a new layer of cubes).

If all of the cubes with "external faces" are painted, there are 2 ways to count them.

The simpler way is "the number of cubes with external faces
is the difference between the total amount of cubes and the amount of internal cubes".
• January 8th 2011, 06:48 AM
Soroban
Hello, abm!

If I read the problem correctly, the question is rather silly.

Besides, some introductory statements are missing:
. . "An $n\!\times\!n\!\times\!n$ cube is painted on all sides,
. . . . . then it is cut up into unit cubes."

Quote:

$\text{For a 10x10x10 cube, 1000 cubes total }(10^3),$
. . $\text{and }8^3\text{ don't have paint on them.}$

$\text{What fraction of the cubes would have a painted side upward?}$

Isn't that simply the top layer of cubes? . $\dfrac{100}{1000} \,=\,\dfrac{1}{10}$

• January 8th 2011, 07:02 AM
abm
Thanks
Thanks for your help. But can you tell me how to calculate internal cubes and external cubes. Suppose we have 3x3x3 cube what will be its internal and external cubes?
• January 8th 2011, 07:11 AM
Quote:

Originally Posted by abm
Thanks for your help. But can you tell me how to calculate internal cubes and external cubes. Suppose we have 3x3x3 cube what will be its internal and external cubes?

Let's say you meant to use the word "outward".

You are in fact given the number of internal cubes.
If the external cubes were removed, you'd have an 8x8x8 cube,
since a layer of cubes would be removed from all 6 faces.

Therefore the number of cubes with external faces are

$10^3-8^3$

If you count them individually, there are 100 on the front and 100 on the back.
There are 80 on the two sides that were not already counted
(since cubes on the front and back have faces on the sides, top and bottom).
Then there are 64 on the top and bottom not already counted.

$200+160+128=488$

$10^3-8^3=1000-512=488$

Be sure that this isn't a trick question referring to the cubes on the top face only
if placed flat on a table for instance.

For a 3x3x3 cube, there is only 1 internal cube.
The external cubes are 9+9+3+3+1+1=26
3x3x3=27.