Slovenija 2008

Denote the digits of $x$ by $a$, $b$ and $c$, so that $x=\overline{abc}$. Three-digit numbers we can form from $a$, $b$ and $c$ are $\overline{abc}$, $\overline{acb}$, $\overline{bac}$, $\overline{bca}$, $\overline{cab}$, $\overline{cba}$ and their sum is $100(2a+2b+2c)+10(2a+2b+2c)+(2a+2b+2c)=222(a+b+c)$. Hence, the sum of the numbers written on the paper is $$3434=222(a+b+c)-\overline{abc}=122a+212b+221c.$$ Consider this equation modulo $9$. When $3434$ is divided by $9$ the remainder is $5$ and the remainder of $122a+212b+221c$ is the same as that of $5a+5b+5c=5(a+b+c)$. We conclude that $a+b+c$ should be congruent to $1$. But $6=1+2+3\le a+b+c\le 7+8+9=24$, so $a+b+c$ can only be $10$ or $19$. If $a+b+c=10$, we have $3434=122a+212b+221c<221(a+b+c)=221\cdot 10<3434$, and this is not possible. So, $a+b+c=19$ and $\overline{abc}=222(a+b+c)-3434=222\cdot 19-3434=784$. The only possible solution is $x=784$.