### A Kindergarten Activity As a Problem for Adolescents

Twelve kids stand in a circle, a kid per one of twelve marked spots. Every now and then one moves clockwise, another counterclockwise to an adjacent position (which may be occupied by more than one kid or be empty.) Is it possible that after a while all of them stand at the same spot?

The movement of the kids can be organized as a random selection. Fill two jars (a "clockwise" jar and a "counterclockwise" jar) with notes or balls carrying the names of the kids. If the same name is drawn from both jars, let the kid sing or jump in place.

It also possible to think of a strategy that would bring all participants to the same spot.

In both cases, it is useful to start with smaller (than 12) numbers. If there are only two players, they will be simply exchanging places with every move. When there are three players, they may converge to a single spot in one move. For four players we again run into a difficulty, but for five it may take just two rounds to get the kids all together. It is practically obvious that there is a strategy for any odd number of kids. For the even number of participants, the situation is clearly more difficult, and the problem seems unsolvable. This is indeed the case. But how that can be proved?

Returning to the situation with 12 kids, it is convenient to think of them as standing on a big clock face. Associate with every kid the hour corresponding to his/her position. Let S be the sum total of all the hours assigned to the participants. At the beginning,

S = 1 + 2 + 3 + ... + 12 = 78

Except when a kid moves from #12 clockwise or #1 counterclockwise, S does not change. In the exceptional cases, S either does not change or changes by 12. It follows that at all times

S = 78 + 12k = 12(k + 6) + 6,

where k is an integer. This means that at all times S ≡ 6 (mod 12). However, if all 12 kids stood on the same spot, S would be divisible by 12 - a contradiction.

Now, the same reasoning applies for any even number N = 2K of participants. In general,

S = K(N + 1) + k·N = N(k + K) + K,

so that S ≡ K = N/2 (mod N). But, when the kids are all at the same spot, S ≡ 0 (mod N).