Slovenian Mathematical Olympiad, 2008

Denote the three-digit number by $\overline{abc}$. Its triple is equal to
$$3 \cdot \overline{abc} = (3a) \cdot 100 + (3b) \cdot 10 + 3c.$$
Obviously, $a$ has to be at least 1. If $a = 1$ and we want the digit at the hundreds in $3 \cdot \overline{abc}$ to be even, we require $3b \cdot 10 + 3c \ge 100$, which implies $10b + c \ge \frac{100}{3} = 33 + \frac{1}{3}$. The least possible number that satisfies this inequality is 34. The solution is 134 and its triple is 402.