Problems & Puzzles: Puzzles Puzzle 236. A Broken Face Break a modern^{(*)} Romannumeral clock face into K pieces such that the sum of the resulting numbers on each piece are distinct prime numbers (p_{1}, p_{2}, ...p_{K}) . Let's call Z to the sum of all these primes, and let's ask for solutions such that Z & K are maximum quantities (All numerals can be read from the center of the clock face.)
Old romannumeral clock face Q1. What are K & Z? (Show the distinct prime numbers thus generated) Q2. Regarding the primes, the solution obtained, is unique, or are there other solutions? _____ Solution: Jon Wharf sent ten solutions. Both came with Z=82 and K=7.
(2, 3, 5, 7, 11, 13, 41)
(2, 3, 5, 7, 11, 17, 37) (2, 3, 5, 7, 11, 23, 31) (2, 3, 5, 11, 13, 17, 31) (2, 3, 5, 7, 13, 23, 29) (2, 3, 5, 7, 17, 19, 29) (2, 3, 5, 11, 13, 19, 29) (2, 3, 7, 11, 13, 17, 29) (2, 3, 5, 13, 17, 19, 23) (2, 3, 7, 11, 17, 19, 23)
All of these can be achieved by breaking the
clock face (always splitting IV and IX). The
partition with max 41 and the two with max 31 can be achieved with each
prime on a continuous section of the circumference. In these cases the max
prime starts in the middle of the IX and uses up the bulk of the X's
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