One of my all-time favorite "Using your Head is Permitted" riddles was the
December 2007 riddle. This month's riddle is a
follow-up to it.
Let n be an integer greater than 1, and let A, B and C be one to one functions, mapping the elements [0..n-1] to themselves.
All three functions satisfy the property that they are cyclic permutations, in the sense that for any x:0≤x<n the list x, A(x), A(A(x)), A(A(A(x))), etc., goes through all n possible elements before returning to x. This is true not only for A, as in the example, but also for B and C.
The question: for which n do such A, B and C exist, where for all x, B(A(x)) = C(x)?
For each n, either provide a proof of impossibility or an example of such an A-B-C triplet.
List of solvers:Christian Blatter (1 June 23:10)
Zilin Jiang (2 June 15:15)
Jin Ruizhang (2 June 23:49)
Oded Margalit (3 June 15:38)
Phil Muhm (4 June 00:25)
Hongcheng Zhu (4 June 17:50)
Lee Siu Hung (7 June 01:53)
Daniel Bitin (7 June 07:19)
Itsik Horovitz (7 June 08:23)
Jan Fricke (21 June 22:42)
Erick Wong (23 June 10:08)
Omer Moussaffi (24 June 06:31)
Albert Stadler (27 June 18:32)
Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.
The solution will be published at the end of the month.
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