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If I have a $10 \times 10$ sided cube (rubik's cube is $3 \times 3$ sided), and dropped it in a bucket of black paint, can you tell me mathematically how I could determine the total number of sides that are black?

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possible duplicate of Breaking a larger cube into a smaller one – Ted Feb 23 '13 at 21:24
Not without the depth of the bucket and the quantity of paint contained therein, with respect to the size of the cube. – Emily Feb 23 '13 at 21:25
@Ted: I don't think this is a duplicate. – Zev Chonoles Feb 23 '13 at 21:26
@Arkamis The mathematical formulation of this is obvious... – Potato Feb 23 '13 at 21:27
Many answers. One could say $6$. Or if it is $1\times 1$ sides of cubelets, $600$. Perhaps they were really asking how many cubelets have at least one black side. That is less immediate. – André Nicolas Feb 23 '13 at 21:29

The obvious answer would be $6\times 10\times 10=600$ since each face of the cube has $10\times 10$ "sides" on it and there are 6 faces of a cube.

However, if this one of those silly "lateral thinking" questions, perhaps the paint seeps between the "cracks" and in fact every "side" of each of the $10\times 10\times 10=1000$ "minicubes" making up a Rubik's-style cube is covered, so that therefore there are $6000$ "sides" painted black.

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+1 for Silly Lateral Thinking. – Inquest Feb 23 '13 at 21:34
In other words: all of 'em! – JohnD Feb 24 '13 at 4:02

The actual question asked MAY have been: if you drop a $10 \times 10 \times 10$ cube into a bucket of black paint, how many "cubies" have AT LEAST one side with black paint on it.

In this case, there is an $8 \times 8 \times 8$ cube in the middle that hasn't been touched. So the answer is $10^3 - 8^3 = 488$.

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