A Puzzle by Midnight Math: April

Imagine you have n points evenly spaced around a circle. Choose one of these as the starting point, p0. Take a pen and draw a line skipping m – 1 points so that you connect p0 to pm.

Continue doing this without lifting the pen (so next you connect pm to the point m points later, and so on). Eventually you get back where you started.

Perhaps you have drawn an n-pointed star (the classic way of drawing a 5 pointed start is with n = 5 and m = 2).

Why can you never get a six pointed start this way? What other values of n will never result in a star?

Send your solutions (with proof) to midnight.math at outlook.com. If you are correct, you will be given the highest of accolades: your name mentioned here, next issue.

Correct answers to last month’s puzzle:
Arash Ushani, Reyner Crosby, Ruby Spring

(see last month’s puzzle and solution at http://franklyspeakingnews.com/node/170)

———————————————————
Arash’s solution:
Well, 2 pointed, 3 pointed, and 4 pointed stars just aren’t exciting, so let’s stick with n > 4.

This is a cyclic group of order n, which is isomorphic to Z_n (the group of integers 0 through n – 1 over addition modulo n). The number of generators for such a group are given by the totient function (ie, the number of integers less than n that are relatively prime to n). For any n > 3, the we always have the generators 1 and n – 1 (corresponding to a n-gon drawn clockwise and counterclockwise). Any other generator leads to a n-pointed star (as it must hit every point, and wouldn’t simply hit them in order, resulting in an n-gon).

For n = 6, the totient function yields n = 2 (1 and 5 are the only generators of Z_6, yielding hexagons). Thus, for n = 6, we have no 6 pointed star.

For no other number n>4 does the totient function of n yield 2. Thus, n=6 is the only arrangement that cannot yield an n-pointed star. For every other n, the totient function yields greater than 2, meaning there is some other generator of Z_n that will yield an n-pointed star.