2. From three different digits, none of which is zero, all possible three-digit numbers are formed. Their sum, increased by 1, is 1999. What are those three digits? List all possible cases.
Solution: Let the digits be $a, b$, and $c$ and let $a<b<c$. All three-digit numbers are: $\overline{a b c}, \overline{a c b}, \overline{b a c}, \overline{b c a}, \overline{c a b}$, and $\overline{c b a}$. We have
$\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}+1=1999$, that is $100 a+10 b+c+100 a+10 c+b+100 b+10 a+c+100 b+10 c+a+100 c+10 a+b+100 c+10 b+a+1=1999 ; 222(a+b+c)=1998$, hence $a+b+c=9$, so $a=1, b=2, c=6 ; a=1$, $b=3, c=5 ; a=2, b=3, c=4$.