A 10 by 10 board has area 100, which 4 divides, so the count alone permits a tiling by 25 pieces of size 1 by 4 laid horizontally or vertically. Does such a tiling exist? Give a colouring argument that settles it.

A 10 by 10 board has area 100, which 4 divides, so the count alone permits a tiling by 25 pieces of size 1 by 4 laid horizontally or vertically. Does such a tiling exist? Give a colouring argument that settles it.

Approach: Find a colouring in which every placement of the piece, in either orientation, covers each colour the same number of times, then count how often each colour occurs on the board.

No tiling exists. Colour the cell in row i and column j, both indices running 0 to 9, with (i + j) mod 4. Any 1 by 4 piece covers four consecutive indices in one coordinate, so in either orientation it covers each of the four colours exactly once. A tiling by 25 pieces would then use each colour exactly 25 times. Counting the board directly, residues 0 and 1 occur 3 times each among the indices 0 to 9 while residues 2 and 3 occur twice, and combining these gives colour class sizes 25, 26, 25 and 24. Those are unequal, so the impossibility follows. The area check that 4 divides 100 settles nothing on its own. The general rule is that an m by n board with 4 dividing mn is tileable by 1 by 4 pieces exactly when 4 divides m or 4 divides n.

Follow-up: Which rectangles can be tiled by 1 by 5 pieces, and what colouring proves the negative cases?

Key concepts: four colouring, counting argument, impossibility proof.