, 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 $x = 784$.