2. A square with a side of 49 cm is given. Divide the given square into 2009 smaller squares of two types, which have integer sides (the side lengths of the divided squares can have one of two different integer values).
Solution. Let's consider squares with sides of $1 \mathrm{~cm}$ and $3 \mathrm{~cm}$. Let the number of squares with a side of $3 \mathrm{~cm}$ be $x$. Then the number of squares with a side of $1 \mathrm{~cm}$ will be 2009 - x. The sum of the areas of the smaller squares will be the same as the sum of the areas of the larger square, i.e.

$$
\begin{aligned}
& (2009-x) \cdot 1 + x \cdot 9 = 49^{2} \\
& 8 x = 392 \\
& x = 49
\end{aligned}
$$

According to this, one of the possible solutions is: 49 squares with a side of $3 \mathrm{~cm}$ and 1960 squares with a side of $1 \mathrm{~cm}$.