Problems & Puzzles: Puzzles Puzzle 569. PrimeMagic Hexagon A few days ago a reader asked to me for the solution for the Magic Hexagon order 3. Instead of solving it I just surfed the web and found the well known solution [after Clifford W. Adams, 1910] that I shared to the reader. This is the asked solution. This is a normal magic solution in the sense that the using the first 19 integers (119) the sum M for the 15 rows is the same numbers (38). After sharing the normal magic solution it was inevitable to start thinking in a possible magic solution using distinct primes. Unfortunately and due to simple parity considerations for sure you too can easily conclude that a magic solution using only distinct primes is impossible. But what if you are able to use the even prime "2" as much as you need in order to satisfy the parity conditions? After some hours of random concentration I came up with a possible scheme: If you use three times distinct powers of the prime "2" in certain positions (for example: instead the position of the 3, 9 & 10 in the normal magic solution) you are able to get a magic solution using odd distinct primes in the rest of the 16 positions.
Q. Using at the most three times distinct powers of 2 (not necessarily the 2^1, 2^2 & 2^3 shown in the sketch above), and distinct 16 odd primes, can you get a magic solution for the Hexagon order 3, with the same sum M in each of the 15 rows?. Of course we are looking for the minimal sum M solution. ________ Contributions came from Seiji Tomita & Carlos Rivera. *** Seiji Tomita wrote:
*** Carlos Rivera wrote:
*** Seiji Tomita wrote (Jan 2011)
*** I pointed out to Mr. Tomita that my M=108 "minimal" solution has a feature a little bit hidden: it is a solution composed by 16 "almost consecutive primes" (it goes from 5 to 67, missing only the prime 61). I asked him to look out for a solution composed by 16 rigorous consecutive primes. This is his answer:
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