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.
Friday, March 14, 2008
Playing with Blocks
Here's a fun puzzle to ponder.
4 comments:
For NPR puzzle posts, don't post the answer or any hints that could lead to the answer before the deadline (usually Thursday at 3pm ET). If you know the answer, submit it to NPR, but don't give it away here.
You may provide indirect hints to the answer to show you know it, but make sure they don't assist with solving. You can openly discuss your hints and the answer after the deadline. Thank you.
Subscribe to:
Post Comments (Atom)
I got it, but I won't post the answer just yet.
ReplyDeleteIf 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.
Anyone else working on this? Or should I just post the answer?
ReplyDeleteThe 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.
ReplyDeleteYou 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.
Answer:
5x5x5 cube with two opposite faces unpainted. 80 cubes with paint, 45 without.
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.
ReplyDelete