3 × 4

complete

smallest rectangle: 3 × 4

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

**Proof.** It suffices to show that it doesn't tile any
(4m + 2) × (4n + 2) rectangle.
Number the squares by

(x, y) |--> { 1 if x and y are both even { 0 otherwise.

No matter how it is placed, each F hexomino covers an even total. However, a (4m + 2) × (4n + 2) rectangle covers an odd total, so it can't be tiled. QED.

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Updated May 25, 2005.