One of my favourite “tiling” puzzles was given in a high school state championship in one of the former Yugoslavia countries in the 90s:

“You have triominos – L-shaped 3-square dominoes. Prove that any 8×8 board with one tile missing anywhere can be covered with triominos”.

The actual problem was slightly different but I feel it’s easier in a sense. It needs a small aha moment.

]]>Grant’s puzzle is very nice, with at least two aha moments. Use 4 colors for 1x1x1 cells inside the box, assigned by (x+y+z) mod 4. Then, a brick occupies one cell of each color regardless of placement. The coloring partitions the cells into sets of sizes 53, 55, 55, 53. So if 53 bricks can be packed, every one of the cells with the 1st and 4th colors must be occupied. Next, the box can be rotated to put any of its 4 long diagonals along the line (0,0,0)-(6,6,6). These rotations generate different colorings. Any packing will leave 4 empty 1x1x1 cells. For any cell you try to leave unoccupied, one of the colorings has that cell in a set of 53, so that cell must be occupied. This contradiction implies there is no such packing.

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