2. Determine all natural numbers that are 33 times larger than the sum of their digits.
Solution. Let the desired number be $\overline{a_{n-1} a_{n-2} \ldots a_{1} a_{0}}$, where $a_{i} \in\{0,1,2, \ldots, 9\}$ for $i=0,1$, ..., $n-1$. The condition

$$
a_{n-1} \cdot 10^{n-1}+10 a_{1}+a_{0}=33\left(a_{n-1}+\ldots+a_{1}+a_{0}\right)
$$

can be written in the form

$$
a_{n-1}\left(10^{n-1}-33\right)+\ldots+a_{2}\left(10^{2}-33\right)=23 a_{1}+32 a_{0}
$$

The left side of the last equation is not less than $10^{n-1}-33$, and the right side is not greater than 495, so this equation is possible only for $n \leq 3$. It is easy to check that there are no one-digit and two-digit numbers that satisfy the given condition. For $n=3$ we get $100 x+10 y+z=33(x+y+z)$. This means that the desired number must be divisible by 3. But this means that the right side is divisible by 9, so the number is divisible by 9. Therefore, $9 \mid x+y+z$, which is possible only if $x+y+z \in\{9,18,27\}$. By direct verification, it is shown that only for $x+y+z=18$ we get the number 594 which satisfies the conditions of the problem.