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!

  • $\begingroup$ I think you can make a little $2\times 2\times 2$ cube, and then a $20\times 20\times 20$ is easy to do. $\endgroup$
    – DiegoMath
    May 30 '14 at 2:08
  • $\begingroup$ @JoelReyesNoche fixed - don't damage any of the brickc. $\endgroup$ May 30 '14 at 2:09
  • $\begingroup$ This raises memories of some happy days, play lots of basket ball, win the national mathematical olympics (1991),... But then the task was to not only stick one needle through it, but some large amount, might have been 100 (simultaneously and in different places). A similar argument to the one above gave a number of over 160, if my memory is right. $\endgroup$ May 30 '14 at 9:55

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.

  • $\begingroup$ I already solved the problem - I was looking for different solutions, if any. $\endgroup$ May 30 '14 at 2:08
  • $\begingroup$ @JoelReyesNoche added solution to OP. $\endgroup$ May 30 '14 at 2:17

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