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You have a 3x3x3 cube, built with 1x1x1 bricks (27 bricks in total).

  1. Aug 30, 2009 #1
    You have a 3x3x3 cube, built with 1x1x1 bricks (27 bricks in total). You remove two diagonally opposite bricks (long diagonals). Can you build the remaining figure from 2xx1x1 bricks?
     
  2. jcsd
  3. Aug 31, 2009 #2
    Re: Bricks

    i'm going to say:
    no
    the shapes volume is now 25 bricks
    this would take 25 1x1x1 bricks
    or 12.5 2x1x1 bricks
    unless your allowed to have half a 2x1x1 brick, you cant do it?
    unless i've missed something
     
  4. Sep 1, 2009 #3
    Re: Bricks

    Yeah, this problem becomes trivial if you've got an odd-dimension cube. However, it may still be impossible if it's a 4x4x4 cube, for different reasons, which I think is what the OP was going for. Just have to imply a different meaning of "Long Diagonal"-- that is, instead, you want to remove 2 cubes diagonally opposite on the same "plane" of the cube.

    DaveE
     
  5. Sep 1, 2009 #4
    Re: Bricks

    Dave is right, I didn't realize the triviality of an odd-order cube :redface:

    Still though, like he pointed out, the 4x4x4 case, or any order for that matter, is impossible too. But the solution (at least the one I know) is much harder!
     
  6. Sep 9, 2009 #5
    Re: Bricks

    There is a problem in dimension with its first digit. So, i am not guessing answer about this bricks game.
     
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