Slovenija 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.