15. Around a table, there are 10 people, each of whom can be either a knight or a knave. Every time a knight speaks, the statement they make is true; every time a knave speaks, the statement they make is false. One of them makes the following statement: "to my right sits a knight and to my left sits a knave." The person to the right of this person states: "to my left sits a knight and to my right sits a knave." The person to the right of this one claims: "to my right sits a knight and to my left sits a knave." And so on, the statements alternate, until the tenth person, who states: "to my left sits a knight and to my right sits a knave." It can be concluded that, among the 10 people present, the total number of knights...

(A) could be 0, 2, 4, 6, or even 8, but not 10

(B) could be 0, 2, or 4, but not 6, 8, or 10

(C) could be 0, 2, 4, or even 6, but not 8 or 10

(D) could be 2, 4, or 6, but not 0, 8, or 10

(E) is definitely 5
(15) The correct answer is $(B)$.

Let's denote the 10 people with the integers from 1 to 10, where 1 is the person who spoke first, 2 is the person who spoke second, and so on.

Person 1 states that 2 is a knight, and 2 says the same about 1: this means that if one of them is a knight, then both are. Therefore, the pair 1, 2 is either two knights or two knaves. The same applies to the pairs 3, 4, 5, 6, 7, 8, and 9, 10. It can also be observed that if one of these pairs is made up of two knights, then the two adjacent pairs must both be made up of knaves, since each of the two knights claims that their neighbor outside the pair is a knave.

Conversely, the pairs adjacent to a pair of knaves can be made up of either knights or knaves (the statements of the aforementioned pair are in any case false for having declared that the other member of the pair of knaves is a knight).

It is therefore clear that either all 10 people are knaves (in which case, indeed, all statements would be false), or there could be 1 or at most 2 pairs of knights, but no more: if there were 3 pairs of knights, at least 2 of these pairs would have to sit next to each other, which is not permissible for the reasons stated above. 

Question proposed by Andrea Ciprietti and Maria Chiara Ricciuti.