Problems & Puzzles: Puzzles

Puzzle 569.- Prime-Magic 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 (1-19) the sum M for the 15 rows is the same numbers (38).

More references: 1, 2, 3

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.

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Warning: Please notice that I'm not totally sure that this proposed scheme will admit one prime solution, or supply the minimal asked sum. Please feel you free to change to other powers of 2, or to change to a totally different approach, if necessary.

Contributions came from Seiji Tomita & Carlos Rivera.

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Seiji Tomita wrote:

M=200:{2,4,8}

2  67 131
109 19 11  61
89  7  17 79  8
107 37 13  43
4  47 149

M=136:{4,8,16}

4 53 79
71 11 13 41
61 5  7 47 16
67 29 17 23
8 31 97

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Carlos Rivera wrote:

My best solution was looking for the minimal solution using any kind of even numbers in the suggested positions was:

M=108:{8,12,18}

8 53 47
41 11 13 43
59  7   5 19 18
37 31 17 23
12 29 67

Is this the real minimal  solution, using three even numbers???

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Seiji Tomita wrote (Jan 2011)

M=108:{4,10,12}
4 61 43
37 11  7 53
67  5  1 23  12
31 47 13 17
10 19 79

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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:

There is no solution using 16 rigorous concecutive primes.

Search condition:
Value of three even numbers < 100
Value of cell < 100

But, there are three more solutions using 16 almost consecutive primes as following.

M=114:{24,30,34}
24 47 43
59 11  7 37
31  3  5 41 34
53 29 13 19
30 23 61

M=138:{38,42,48}
38 71 29
47 11 19 61
53 13 7 17 48
43 41 23 31
42 37 59

M=154:{64,70,74}
64 71 19
59 23 11 61
31 7 13 29 74
53 41 17 43
70 47 37

It seems that M=108 is a minimal solution.

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