2. From the set $\{10 ; 11 ; 12 ; \ldots ; 19\}$, 5 different numbers were chosen, and from the set $\{90 ; 91 ; 92 ; \ldots ; 99\}$, 5 different numbers were also chosen. It turned out that the difference between any two of the ten chosen numbers is not divisible by 10. Find the sum of all 10 chosen numbers.
Solution. Obviously, all the selected numbers have different last digits. Therefore, the sum can be found by separately adding the tens and units of these numbers. Among the tens, there are five 10s and five 90s, so the sum of the tens is $5 \cdot 10 + 5 \cdot 90 = 500$. And among the units, each digit appears exactly once, so their sum is: $0 + 1 + 2 + \ldots + 9 = 45$. Therefore, the sum of all ten numbers is 545.

Answer: 545.