2. There are 81 squares of the same size. Arrange them into two rectangles so that their perimeters differ by a factor of 2. No squares should be left over.

(L. S. Korechkova)
Answer: $3 \times 19$ and $3 \times 8$ or $3 \times 23$ and $1 \times 12$.

Solution. The sum of their areas is 81 ( $57+24$ in the first option and $69+12$ in the second), and their perimeters indeed differ by a factor of 2: $2 \cdot(3+8)=22,2 \cdot(3+19)=44$ or $2 \cdot(1+12)=26,2 \cdot(3+23)=52$.

Criteria. A correct example is provided without calculations showing that it fits - 5 points. If areas and perimeters are confused in the solution - 0 points.