2. If you take three different digits, form all six possible two-digit numbers using two different digits, and add these numbers, the result is 462. Find these digits. Provide all variants and prove that there are no others.
Solution. Let these three digits be denoted by $a, b, c$. We obtain six numbers $10 a+b, 10 b+ a, 10 c+b, 10 b+c, 10 a+c, 10 c+a$. The sum of these numbers is $22(a+b+c)$. According to the condition, $a+b+c=462: 22=21$. By trial, we find the sets of digits $(6,7,8),(4,8,9),(5,7,9)$. These sets can be obtained, for example, by enumeration based on the largest digit. Let's choose the digit 9. The sum of the two remaining digits is 12. Taking into account that the digits are different, we get the sets $(4,8,9),(5,7,9)$. If we decrease the largest digit by 1, the sum of the other two digits will become 13. We get the only such set $(6,7,8)$. Reducing the largest digit to 7 is no longer possible.

Answer. $(6,7,8),(4,8,9),(5,7,9)$