Problem:
There are mathematicians seated around a circular table with seats numbered in clockwise order. After a break they again sit around the table. The mathematicians note that there is a positive integer such that
for each , the mathematician who was seated in seat before the break is seated in seat after the break (where seat is seat );
for every pair of mathematicians, the number of mathematicians sitting between them after the break, counting in both the clockwise and the counterclockwise directions, is different from either of the number of mathematicians sitting between them before the break.
Find the number of possible values of with .
Solution:
If , then the mathematician seated in seat before the break will be seated in seat , which is seat , after the break. This is impossible because a mathematician seated in seat before the break will also be seated in seat after the break, violating condition . On the other hand, if , then is a complete set of residue classes modulo , and hence condition is satisfied.
Condition holds if and only if for all and with , and , that is, and . Thus both and are relatively prime to .
Combining the above, a pair of positive integers satisfies the conditions of the problem if and only if
Call such a pair a good pair.
Note that divides for every positive integer . Hence if , then is not a good pair. On the other hand, if , then is a good pair. Therefore all integers with and satisfy the conditions of the problem. These are exactly the integers of form either or with , and there are such integers.
The problems on this page are the property of the MAA's American Mathematics Competitions