Interesting 3x3 table, to form a numbers of unique prime numbers

[jasen]

 

 151  131 359 131  151
    \   ^   ^   ^ /
  131 < 1   3   1 > 131
  353 < 3   5   3 > 353
  191 < 1   9   1 > 191
    /   V   V   V \  
  151  131 953 131  151

This 3x3 table have an interesting properties. 
Every direction (up,down, Right, Left, 45°,135°,225°,315°) of 3 cell form prime numbers.
There are 6 unique prime numbers from this table, they are 131, 151, 191, 353, 359, 953.

Create a more interesting 3x3 table with the same rule, which there are 9 unique prime numbers from the table.

[tojo928]

my bad. I left a couple options for the exterior primes off my list.

On ‎11‎/‎3‎/‎2016 at 9:52 PM, jasen said:

It is possible, I have it. I will show it next week if nobody find it until 10 Nov 2016.

 

Spoiler

3 1 1

1 5 7

1 7 9

 

Fun extra question to add in to challenge on this one. What is the lowest number of unique primes that could be used in this and how many unique ways (including rotation and reflection) can this be made?

[bonanova]

First thoughts

Spoiler

General table gives 16 numbers:
abc def ghi cba fed ihg adg beh cfi gda heb ifc aei iea gec ceg

a b c
d e f
g h i

Explore ways to reduce this to nine numbers:

Try making all the corners the same:

a b a
d e f
a h a

This in fact gives nine numbers: aba def aha fed ada beh afa heb and aea.

We can't reduce this further: no other position can be a because 3 always divides aaa; the center e can't be an edge number without e.g. making aba = aea; and no two edge numbers can be the same without e.g.making aba = afa. In both cases there are only eight distinct numbers.

So if the corners are equal, then a b c f and h must be distinct. But that is impossible:

Edge numbers (i.e., all numbers other than e) must all be members of {1 3 7 9}. A number that ends in any another digit is even or a multiple of 5. The values of a b d f and h cannot be taken from {1 3 7 9} and also be distinct. So our starting point is wrong: 

The four corner numbers cannot all be the same.

Anyway, back to the puzzle:

Primes must end in {1 3 7 9} and in this puzzle they must also begin with them. These primes are as follows, sorted by leading digit: (palindromes highlighted in red):

P1: 101 107 113 131 151 157 167 181 191
P3: 311 313 337 347 353 383
P7: 701 709 727 733 739 743 751 757 761 787 797
P9: 907 919 929 937

Since only e does not end a number, the numbers whose second digit is even or 5 can only pass through the center of the table. In this list 0 2 4 6 8 and 5 are all present as a second digit, as follows:

C0: 101 107 701 709 907
C2: 727 929
C4: 347 743
C5: 151 157 353 751 757
C6: 167 761
C8: 181 383 787

More later ...

 

[tojo928]

Spoiler

Not possible to get 9 unique values

 

Most you can get is 8 with this(and all the unique rotations and reflections):

7   3  9

3 8/5 3

3   3  7

 

 

 

[jasen]

It is possible, I have it. I will show it next week if nobody find it until 10 Nov 2016.

6 hours ago, tojo928 said:

  Hide contents

Not possible to get 9 unique values

 

 

[jasen]

On 11/6/2016 at 2:39 PM, tojo928 said:

Fun extra question to add in to challenge on this one. What is the lowest number of unique primes that could be used in this and how many unique ways (including rotation and reflection) can this be made?

Spoiler

 

3 unique primes, there are some solution for 3 unique primes, one of them is 

1 3 1     

3 5 3

1 3 1