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

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?
 i'm going to say: Spoiler 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
 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

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

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

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!