2. Find all trimorphic numbers that are equal to 30 times the sum of their digits.
II/2. We are looking for such numbers $\overline{x y z}$ for which $100 x+10 y+z=30(x+y+z)$ holds. Since a 30-fold number must end with the digit 0, we have $z=0$. Then $100 x+10 y=30(x+y)$, or $10 x+y=3(x+y)$. From this, we get $7 x=2 y$. Since $x$ and $y$ are digits, the only possibility is $x=2$ and $y=7$. There is only one three-digit number that is equal to 30 times the sum of its digits, which is the number 270.

Writing the equation $100 x+10 y+z=30(x+y+z)$ or an equivalent: 2 points. The realization that $z=0: \mathbf{1}$ point. The conclusion that $x=2$ and $y=7: \mathbf{2}+\mathbf{2}$ points.