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Iterated Specs (Posted on 2010-05-28) Difficulty: 2 of 5
There is a number N such that if you inscribe an N-gon in a circle, with all its angles an equal number of integral degrees, that N-gon will necessarily be a regular N-gon, with equal side lengths.

Also, if M is the number of possible divisors of N, including 1 and N itself, then the number M also meets the same criteria: If you inscribe an M-gon in a circle, with all its angles an equal number of integral degrees, that M-gon will necessarily be a regular M-gon, with equal side lengths.

What are N and M?

See The Solution Submitted by Charlie    
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Some Thoughts More answers? (SPOILER?) | Comment 3 of 4 |
I don't see any reason that N and M need to be odd.  All we need is N and M to both divide 360 evenly, and be greater than 2.

In addition to (9,3), I think that (6,4), (10,4), (12,6), and (18,6) all do it.  Also (20,6), (24,8), (36,9), (90,12), (180,18), (360,24), and possibly others that I have missed.

Am I misunderstanding the problem?  Have I miscalculated (all done by hand)

  Posted by Steve Herman on 2010-05-28 16:06:03
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