3 × 4

complete

smallest rectangle: 3 × 4

**Proposition.** Any rectangle tiled by the J hexomino has one side
divisible by 4.

**Proof.** It suffices to show that it doesn't tile any
(4m + 2) × (4n + 2) rectangle.
Note that they must mate in pairs

1 2 2 2 1 1 1 1 2 1 2 or 1 2 1 1 1 1 2 2 1 2 2 2 2

so we need only consider tilings by these two shapes. Now consider the numbering

(x, y) |---> { 1 if x or y , but not both, are divisible by 4 { 0 otherwise.

No matter how they are placed, either of these shapes covers an odd total. A (4m + 2) × (4n + 2) rectangle covers an even total. However, it would be tiled by an odd number of the two shapes above, which would cover an odd total, a contradiction. QED.

**Remark.** The J hexomino has a signed tiling of a 2 × 3
rectangle.

. . . * * * * * * . . . . . * * . . * . * . . . * . * + * * . - * . . = * * * * * * . . . . . . * * *

Problem 3 from the 2004 International Mathematical Olympiad asks which rectangles can be tiled by this hexomino. The proposition above is key to the solution.

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Updated August 23, 2011.