2. There are 49 identical squares. Form two rectangles from them so that their perimeters differ by a factor of 2. No squares should be left over.

(L. S. Koreschkova)
Answer: $7 \times 4$ and $1 \times 21$ or $1 \times 33$ and $1 \times 16$.

Solution. The sum of their areas is 49 ($28+21$ in the first case and $33+16$ in the second), and their perimeters indeed differ by a factor of 2: in the first case $2 \cdot(7+4)=22$, $2 \cdot(1+21)=44$; in the second $2 \cdot(1+16)=34, 2 \cdot(1+33)=68$.

Criteria. A correct example is provided without calculations showing that it fits - 5 points. At least one of the sides is not an integer - 0 points.