Problem 3

Define a hook to be a figure made up of six unit squares consisting of a row of three squares and a column of four squares sharing a corner square, together with one further square attached at the far end of the row (or any figure obtained from this one by rotations and reflections). Determine all

m\times n

rectangles that can be covered without gaps and without overlaps by such hooks, with no part of a hook covering area outside the rectangle.

Step 6 of 6: Conclude both divisibility conditions

Previous step Next step

\text{applying Step 5 to both }6a\times2b\text{-type sub-rectangles} \implies 3\mid m\text{ or }3\mid n,\quad 4\mid m\text{ or }4\mid n

Detailed analysis

The pairing gives $12\mid mn$ . Hence $3$ divides at least one of $m,n$ . It remains to prove that a side is divisible by $4$ . Assume for contradiction that $4\nmid m$ and $4\nmid n$ . Since $4\mid mn$ , both sides are even; after interchanging them if necessary, the side divisible by $3$ is $m=6a$ and the other is $n=2b$ . The assumptions $4\nmid m,n$ say that $a,b$ are odd. Thus the number of 12-cell tiles is $mn/12=ab$ , which is odd. But Step 5, and the same argument with rows and columns exchanged, show that the numbers of both tile types are even, so their total is even, a contradiction. Therefore $4\mid m$ or $4\mid n$ , and together with the area conclusion this proves $3\mid m$ or $3\mid n$ and $4\mid m$ or $4\mid n$ .