1. Jaka thought of a trimest number $x$, which in $\mathrm{v}$ notation has different non-zero digits. Then he wrote down all other trimest numbers that he could write using the digits of the number $x$. Determine all possible numbers $x$, if the sum of the numbers on the list is 3434.
III/1. Let's denote the digits of the number $x$ with $a, b$, and $c$, so $x=\overline{a b c}$. All the three-digit numbers formed by the digits $a, b$, and $c$ are $\overline{a b c}, \overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}, \overline{c b a}$, and their sum is $100(2 a+2 b+2 c)+10(2 a+2 b+2 c)+(2 a+2 b+2 c)=222(a+b+c)$. The sum of the numbers on the list is thus

$$
3434=222(a+b+c)-\overline{a b c}=122 a+212 b+221 c
$$

Let's consider the above equation in terms of divisibility by 9. The remainder of the number 3434 when divided by 9 is 5, and the remainder of the number $122 a+212 b+221 c$ is the same as the remainder of the number $5 a+5 b+5 c=5(a+b+c)$. From this, it follows that the remainder of the number $a+b+c$ when divided by 9 must be 1. Since $6=1+2+3 \leq a+b+c \leq 7+8+9=24$, $a+b+c$ must be either 10 or 19.

If $a+b+c=10$, we can estimate $3434=122 a+212 b+221 c<221(a+b+c)=$ $221 \cdot 10<3434$, which is not possible. Therefore, $a+b+c=19$. Now we can calculate $\overline{a b c}=$ $222(a+b+c)-3434=222 \cdot 19-3434=784$. The only possibility is $x=784$.[^2]