3. Once, 20 animals gathered in the jungle, 11 of them were tigers, who always tell the truth, and the remaining 9 were monkeys, who always lie. Around a stump, $n \geq 3$ animals sat, each of them stating, "Exactly one of my two neighbors is a monkey." For which $n$ is this possible? (It is necessary to find all possible $n$ and explain why others cannot be.)
Answer. $n=3,4,5,6,7,8,9,12,15$.

Solution. First, there is a case when only monkeys sit behind the stump: in this case, all $n \leq 9$ are suitable.

Now let there be at least one tiger behind the stump. Let this tiger be $A$, its neighbors be tiger $B$ and monkey $C$. Then on the other side of $B$ sits a monkey, and on the other side of $C$ - a tiger. Continuing this reasoning, we get that those sitting behind the stump are divided into non-overlapping triples, each of which has two tigers and one monkey. Therefore, $n=3k$, and behind the stump there are $k$ monkeys and $2k$ tigers. From the condition $k \leq 9$ and $2k \leq 11$ it follows that $k \leq 5$. Thus, two more possible values are added: $n=12, n=15$.

Criteria. Full solution - 7 points. For losing one case, deduct 2 points. If 3 or more cases are lost - 0 points. All variants are found, but there is no proof that there can be no others - 3 points.