Ken's POTW


Primes on a Clock
  1. Leave 6 adjacent numbers of the face of a clock intact and rearrange the other 6 in such a way that the sum of every pair of adjacent numbers is prime.
  2. Leave as many numbers as possible in their correct positions on the face of a clock and rearrange the rest such that the sum of every pair of adjacent numbers is prime.
  3. Repeat the previous problem, such that the sum of every adjacent pair is composite (not prime).

Source: A small book called Problematical Recreations shown to me by Bill Cleveland. Used as Ken's POTD 3/7/94.


Solutions were received from Nick Baxter, Kirk Bresniker, Rich Polster, Bill Beall, Philippe Fondanaiche, Bert Sevenhant, Louis Elrod.
Nick Baxter's solutions (which are representative of all of them):
1.Leave 6 adjacent numbers of the face of a clock intact 
        and rearrange the other 6 in such a way that the 
        sum of every pair of adjacent numbers is prime.

two solutions:
        1 2 3 4 7 10 9 8 5 6 11 12
        1 2 3 4 9 10 7 6 5 8 11 12

2.Leave as many numbers as possible in their correct 
        positions on the face of a clock and rearrange the 
        rest such that the sum of every pair of adjacent 
        numbers is prime.

move 4 (actually done with 2 swaps):
        11 2 3 8 5 6 7 4 9 10 1 12 

The standard clock has 3 locations (4 5)(7 8)(10 11) where
there is a composite sum.  At least one of each of these
pairs must be moved.  The 4 choices are to cycle either the
even or odd values - none of which give all primes.  So 4
moves must be the minimum.

3.Repeat the previous problem, such that the sum of every 
        adjacent pair is composite (not prime). 

Similar to analysis for #2, if 5 were the minimum number
of moves, then 6,9 and one from each of (11 12)(1 2)(3 4).
The 4 ways to select that also include one from each
pair (12,1) and (2,3) are (11 1 3)(12 1 3)(12 2 3)(12 2 4),
and none produce a solution.

For 6 moves, assume that 1-3,6,9,12 are to move.  Only 1
goes in the 6 position and 2 goes in the 9 position.  There
are 4 ways to permute the remaining, one of which leads to:
        6 9 12 4 5 1 7 8 2 10 11 3

Kirk Bresniker found all possible solutions.
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