The digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 were written (in an unknown order) around a circle. Reading these digits in groups of three in a clockwise direction, nine three-digit numbers are formed. Determine the sum of these nine numbers.
In the following figure, we represent the nine digits written around the circumference.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-088.jpg?height=522&width=534&top_left_y=407&top_left_x=795)

Reading the digits written around the circumference, in groups of three, in a clockwise direction, we obtain the following three-digit numbers:

$$
a_{1} a_{2} a_{3}, a_{2} a_{3} a_{4}, a_{3} a_{4} a_{5}, a_{4} a_{5} a_{6}, a_{5} a_{6} a_{7}, a_{6} a_{7} a_{8}, a_{7} a_{8} a_{9}, a_{8} a_{9} a_{1} \text { and } a_{9} a_{1} a_{2}
$$

To add these numbers, we use the addition algorithm, as indicated below.

$$
\begin{array}{r}
a_{1} a_{2} a_{3} \\
a_{2} a_{3} a_{4} \\
a_{3} a_{4} a_{5} \\
a_{4} a_{5} a_{6} \\
+\quad a_{5} a_{6} a_{7} \\
a_{6} a_{7} a_{8} \\
a_{7} a_{8} a_{9} \\
a_{8} a_{9} a_{1} \\
a_{9} a_{1} a_{2} \\
\hline ? ? ? ? ? ? ?
\end{array}
$$

Analyzing these nine numbers, we notice that all of them have different units digits. Therefore,

$$
a_{3}+a_{4}+a_{5}+a_{6}+a_{7}+a_{8}+a_{9}+a_{1}+a_{2}=1+2+3+4+5+6+7+8+9=45
$$

Similarly, these nine numbers also have all different tens digits and all different hundreds digits. Therefore, the sum of the tens digits is also 45, and the same applies to the hundreds digits. Thus, to add them, we simply calculate \(45 + 45 \times 10 + 45 \times 100 = 4995\). Therefore, the sum of the nine numbers is 4995.