Exercise 1. Seven people dine together every Saturday evening around a round table.

How many times is it possible to dine if everyone wants to have two new neighbors each time?

What is the result for eight people?
Solution: Under the conditions stated, at each dinner, a given person has two new neighbors. Therefore, after the second dinner, they will have had four different neighbors, six different neighbors after the third, and eight different neighbors after the fourth, which is not possible: if there are seven people in total, each can only have six different neighbors, which limits the number of dinners that can be organized under the stated conditions to three. The result is the same if there are eight people in total.

It remains to prove that three dinners can be organized so that each person has two new neighbors at each dinner. At the first dinner, the seven people are numbered in order from 1 to 7: 1234567. At the second dinner, compared to the first, they are placed "one person apart": 1357246, and at the third, "one person apart every three": 1473625. It is easy to verify that such a placement works, but it is not the only one.

If an eighth person is added, it is sufficient to insert them so that they also have two new neighbors each time, for example: 12834567 then 13587246 and finally 18473625. Thus, the following pairs will never have been neighbors: 2 and 3, 5 and 7, 1 and 4, as well as 8 and 6.