BMO2 2002 Q2


By Peter Conlon on Wednesday, February 27, 2002 - 09:05 am:

2. A conference hall has a round table with n chairs. There are n delegates to the conference. The first delegate chooses his or her seat arbitrarily. Thereafter the (k+1)th delegate sits k places to the right of the kth delegate, for 1kn-1. (In particular, the second delegate sits next to the first.) No chair can be occupied by more than one delegate.

Find the set of values n for which this is possible.


By Peter Conlon on Thursday, February 28, 2002 - 04:35 pm:

For question two, my answer was that n must be a power of 2. This wasn't proved very well at all. I treated each delegate as a triangle number (0,3,6,10,15etc), and tried to show that the triangle numbers up to the nth triangle number were all different (mod n) only for n = 2r . I'm not even sure this is right, but I can prove n can't be odd.


By David Loeffler on Thursday, February 28, 2002 - 05:18 pm:

You seem to have the right idea for Q2; the way I did it was to consider factorising n as 2a (2b+1) for integers a and b, and showing that if b > 0 there always exist j and k with j(j+1)/2 congruent to k(k+1)/2 mod n. (In fact I think you can always make j(j+1)/2 - k(k+1)/2 = n; take two cases according to whether or not b > 2a ).

David