1. Determine the number $n$, if it satisfies the following three conditions:

a) all its digits are different

b) from its digits, six different two-digit numbers can be formed

c) the sum of those two-digit numbers is $2 n$.
Solution. From condition b), we obtain that the number $n$ is a three-digit number. Let $n=\overline{a b c}$ where $a, b, c$ are distinct digits. Then the sum of the numbers $\overline{a b}, \overline{b c}, \overline{a c}, \overline{c a}, \overline{c b}, \overline{b a}$ is $2 n=2 \overline{a b c}$, i.e.,

$$
\overline{a b}+\overline{b c}+\overline{a c}+\overline{c a}+\overline{c b}+\overline{b a}=2 \cdot \overline{a b c}
$$

According to this,

$$
22(a+b+c)=200 a+20 b+2 c \Rightarrow 178 a-2 b-20 c=0 \Rightarrow 89 a-b-10 c=0
$$

We can rewrite the last equality as $a+b=10(9 a-c)$. Since $a+b \leq 18$, we get $9 a-c<2$, i.e., $9 a-c=1$. From the equality $9 a=1+c$, we obtain $a=1$ and $c=8$. Finally, $b=9$ and the desired number is 198.