1A. Dimitar arranged the digits $1,2,3,4,5,6,7,8,9$ in a circle in some way. Any three consecutive digits, in the clockwise direction, form a three-digit number. Dimitar added all such numbers. What sum did Dimitar get?
Solution. Each recorded digit will appear in three different numbers, once as a hundreds digit, once as a tens digit, and once as a units digit. Therefore, the digit $n$ will contribute to the total sum by

$$
(1+10+100) n=111 n
$$

Since there are 9 such numbers in total, we get that the total sum will be

$$
111(1+2+3+4+5+6+7+8+9)=111 \cdot \frac{9 \cdot 10}{2}=111 \cdot 45=4995
$$