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You’ve got a 10 x 10 x 10 cube made up of 1 x 1 x 1 smaller cubes. The outside of the larger cube is completely painted red. On how many of the smaller cubes is there any red paint?

The easiest way for me to answer this is this way: There are 8*8*8 cubes not painted = 512 cubes not painted. So 1-512 = 488 cubes painted. Done.

2nd way: there are 2 10*10 faces that are painted + 2 10*8 faces that are painted + 2 8*8 faces that are painted = 2(100) + 2(80) + 2(64) = 200 + 160 + 128 = 360 + 128 = 488

3rd way that I'm having trouble with: I'm trying to add up the outside surfaces together and then subtract the ones that I've double counted. How do I do this? 10*10*6 = 600 600 - double counted ones. 600 - (top part of the cube that was double counted + cubes along the height + cubes on the bottom)

600 - (36 + 32 + 36) = 104 600 - 104 = 496.

Aragh. What am I doing wrong in the third method?

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The angle cubes are counted three times so you need to substract eight –  user68061 Jan 28 at 18:30
    
You are counting every edge three times –  user127.0.0.1 Jan 28 at 18:30
    
Let me think about this... I think you're right. I've only subtracted out the cubes that were counted twice. –  Jwan622 Jan 28 at 18:36

1 Answer 1

For the third method you need to do faces (1 side painted) - edges (2 sides painted) + vertices (3 sides painted). The idea is that the edges get counted twice by each face so we fix by subtracting that total back out. But then the vertices (corners of the cube) end up not being counted at all since each vertex is hit by 3 faces and 3 edges have we have just done 3 - 3. So we have to add back in the number of vertices.

There are 6 faces with 10*10 cubes which gives 600. There are 12 edges with 10 cubes which gives 120. There are 8 vertices (corners) with a single cube each which gives 8. So we have $600-120+8 = 488$.

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