A 20 × 20 × 20 cube is built of 1 × 2 × 2 bricks. Prove that one can pierce it by a needle without piercing a brick.
Taken from Engel's book, but no solution was given.
Here's my solution:
Look down one of the faces of the cube. Call the 19^2 lines through which a needle may potentially be pierced "axes." Now assume there is no path for the needle to pierce without piercing a brick.
(1) each axis must have at least 1 2x2 blocking it - else a needle could pass through.
(2) each column must have an even number of 2x2's
Using (1) and (2) we see that the axis with coords (1,1) must have at least 2 2x2's (if it had only 1, the corner column would have an odd number of 2x2). Then (1,2) and (2,1) must also have >= 2 2x2's. In this way the entire outermost square of axes must have 2 2x2's.
Move in to the next outermost ring of axes i.e. the axes with coordinares (2, 2) (2, 3)... By the same logic, these axes must have >= 2 2x2's. In this way we see that every axis has to have >= 2 2x2's, or over 666 2x2's for each pair of faces. For all three pairs of faces, over 2000 total bricks are needed. But a 20x20x20 cube can hold only 2000 bricks. Contradiction!
Hint: how many places can you insert a needle?
Hint: how many small blocks are there?
Hint: show that if a needle would pierce one block, then it must pierce an even number of blocks.