Friday, March 14, 2008

Playing with Blocks

Here's a fun puzzle to ponder.
A certain number of faces of a large wooden cube are stained. Then the block is divided into equal-sized smaller cubes. Counting we find that there are exactly 45 smaller cubes that are unstained. How many faces of the big cube were originally stained?
Feel free to add a comment with your answer, along with how you solved it.


  1. I got it, but I won't post the answer just yet.

    If you make "m" cuts in each direction, then each face will be divided a grid (m+1) by (m+1).

    There will be (m+1)^3 small cubes, of which (m-1)^3 will be internal cubes and must be unpainted.

    Based on this information, there are only 2 possible values for "m" such that 45 cubes remain unpainted.

  2. Anyone else working on this? Or should I just post the answer?

  3. The number of cubes on a side has to be between 4 and 5. Any less and you don't have enough smaller cubes. Any more and you have too many unpainted cubes in the middle.

    You can try to use a 4x4x4 (which has 8 unpainted cubes on the interior), but you won't find a way to paint some of the sides to leave 37 fully unpainted cubes.

    With a 5x5x5 cube, you have 27 in the interior and you need 18 more. The way to do that is to leave two ends unpainted. That will result in 9 more on each face that are fully unpainted.

    5x5x5 cube with two opposite faces unpainted. 80 cubes with paint, 45 without.

  4. Another approach: in order to attain 45 unstained blocks, the obvious dimension of the region of their source is a 3 x 3 x 5 block segment. since the min dimension which would have a 3 x 3 interior of a stained perimeter would be 5 x 5, and the height of the proposed segment would be 5, the initial block must be 5 x 5 x 5 and in order to be 5 high, both ends must be unstained.