2. Determine in how many ways the vertices of a regular nonagon $A B C D E F G H I$ can be assigned numbers from the set $\{17,27,37,47,57,67,77,87,97\}$ such that each number is assigned to a different vertex and the sum of the numbers assigned to each set of three consecutive vertices is divisible by three.
2. First, let us realize that the numbers 27, 57, and 87 are divisible by three, the numbers 37, 67, and 97 give a remainder of 1 when divided by three, and finally, the numbers 17, 47, and 77 give a remainder of 2 when divided by three. Let us denote $\overline{0}=\{27,57,87\}, \overline{1}=\{37,67,97\}$, and $\overline{2}=\{17,47,77\}$. Now consider a regular nonagon $A B C D E F G H I$. When examining the divisibility by three of the sums of triples of natural numbers assigned to three consecutive vertices of the considered nonagon, it suffices to consider only the remainders of the given numbers when divided by three. Three numbers can give a sum divisible by three only if all three give the same remainder when divided by three (belong to the same residue class), or if they belong to different residue classes. However, if three consecutive vertices were from the same residue class, the number of the next vertex (in either direction) would also have to be from the same class, and thus all subsequent numbers would be from the same class. We do not have such nine numbers, so for any three consecutive vertices, the numbers must belong to different residue classes.

First, assume that the vertex $A$ is assigned a number from the set $\overline{0}$. In this case, the vertices of the considered nonagon can be assigned the residue classes $\overline{0}, \overline{1}, \overline{2}$ of the numbers we are assigning in only two ways, depending on which of the two neighboring vertices of $A$ is assigned $\overline{1}$ and which $\overline{2}$. The residue classes of the other vertices are then uniquely determined with respect to the divisibility condition. The resulting assignment can be written as

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{0}, \overline{1}, \overline{2}, \overline{0}, \overline{1}, \overline{2}, \overline{0}, \overline{1}, \overline{2}),
$$

or

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{0}, \overline{2}, \overline{1}, \overline{0}, \overline{2}, \overline{1}, \overline{0}, \overline{2}, \overline{1})
$$

If the vertex $A$ is similarly assigned the residue class $\overline{1}$, the same applies, either

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{1}, \overline{0}, \overline{2}, \overline{1}, \overline{0}, \overline{2}, \overline{1}, \overline{0}, \overline{2}),
$$

or

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{1}, \overline{2}, \overline{0}, \overline{1}, \overline{2}, \overline{0}, \overline{1}, \overline{2}, \overline{0})
$$

Finally, if the vertex $A$ is assigned the residue class $\overline{2}$, it applies either

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{2}, \overline{0}, \overline{1}, \overline{2}, \overline{0}, \overline{1}, \overline{2}, \overline{0}, \overline{1}),
$$

or

$$
(A, B, C, D, E, F, G, H, I) \leftrightarrow(\overline{2}, \overline{1}, \overline{0}, \overline{2}, \overline{1}, \overline{0}, \overline{2}, \overline{1}, \overline{0}) .
$$

According to the number of ways to sequentially select numbers from the individual residue classes, each of the cases corresponds to

$$
(3 \cdot 2 \cdot 1) \cdot(3 \cdot 2 \cdot 1) \cdot(3 \cdot 2 \cdot 1)=6^{3}
$$

possibilities. Given that no other case can occur besides the six mentioned, we see that the number of ways to assign the given nine numbers to the vertices of the considered nonagon is

$$
3 \cdot\left(2 \cdot 6^{3}\right)=6^{4}=1296
$$

For a complete solution, award 6 points, of which 2 points for justifying that the numbers assigned to three consecutive vertices have different remainders when divided by three. For each omitted case of the six, deduct 1 point. For an incorrectly calculated number of possibilities, deduct 2 points; for a less serious numerical error, deduct 1 point. For a correct calculation of the number of possibilities that the solver considered, even if not all possibilities were listed, award 2 points.