6.6. $a, b, c$ are three different digits. If you add all six two-digit numbers that can be written using them without repeating the same digit twice in one number, the result is 528. Find these digits.
Solution. Let's assume for definiteness $a<b<c$. From these digits, we can form the numbers
$\overline{a b}=10 a+b, \overline{a c}=10 a+c, \overline{b c}=10 b+c, \overline{b a}=10 b+a, \overline{c a}=10 c+a, \overline{c b}=10 c+b$.

Their sum, obviously, gives $22 a+22 b+22 c$, which, according to the condition, equals 528.

From this, $a+b+c=24$. Since $a, b, c$ are different digits, the sum of any two of them must be less than 18 (one digit can be 9, and the other must be less). Therefore, for $a$ we get a possible value of 7, then the sum $b+c=17$. Hence, $b=8, c=9$.

Thus, the required digits are $7,8,9$.

Answer: the digits $7,8,9$