Problems & Puzzles: Puzzles

Puzzle 993. Follow-up to Puzzle 992

Emmanuel Vantieghem sent this nice follow-up to the Puzzle 992:

After I worked  on puzzle 929 I wondered if it is possible to construct a 'decent' dice.
 
Such a dice should contain on every face an NxN matrix of primes in such a way that the matrices on opposite faces sum to a simple matrix consisting of 10's.
 
For instance if the matrix on one of the sides would be
 
   9 2 9
   1 5 7
   9 1 1

then the opposite face should contain the matrix
   
   1 8 1
   9 5 3
   1 9 9
 
Of course, the dice must still contain 12N  different primes, and no zero digit is allowed.
 
At this moment, I'm not able to construct such a dice.  The problem is harder than puzzle 992. But I'm pretty sure that there exist at least one (maybe with an  N  bigger than 3).

Q. Find a decent dice like the asked by Emmanuel for the minimal N possible (we may drop the minimal sum requested in Puzzle 992) and as many other dice you are able to get for larger N values.

 


Contributions came from Paul Cleary, Michael Hürter, Oscar Volpatti, Emmanuel Vantieghem, during the week 14-21 March, 2020:

***

Paul wrote:

I was able to find 1656 decent dice with N = 3.  Here are a few solutions

 

2            2           3

7            5           1

1            1           3

 

8            8           7

3            5           9

9            9           7

 

1            2           7

8            2           7

1            7           3

 

9            8           3

2            8           3

9            3           7

 

6            3           1

6            4           3

1            9           9

 

4            7           9

4            6           7

9            1           1

 

Total 19980  and

 

7            6           1

4            6           7

3            1           3

 

3            4           9

6            4           3

7            9           7

 

1            9           9

8            8           1

1            3           9

 

9            1           1

2            2           9

9            7           1

 

4            9           1

5            4           1

7            7           3

 

6            1           9

5            6           9

3            3           7

 

Also total 19980.  In fact all 1656 solutions have the same total of 19980.

***

Michael wrote:

I found solutions for n = 3, 5, 6, 7, 8, 9 and 10. I did not print the opposite faces.

 

n = 3:

 

face 0
827
229
937
face 1
661
139
911
face 2
227
757
113

 

n = 5:

 

face 0
98899
71713
79423
14341
13339
face 1
33547
86413
87517
39877
33391
face 2
63199
57223
22381
84793
73939

 

n = 6:

face 0
674563
935353
131783
318559
347887
373393
face 1
594827
135461
572269
147293
483629
911773
face 2
863197
563881
671339
364943
695467
911917

 

n = 7:

face 0
6651521
4966151
3545483
8574977
5665949
8434427
1377317
face 1
7438997
7285697
6389777
4395857
1393121
3437363
1771139
face 2
4585433
6566597
4167491
6755699
3124193
5758913
7339373

 

n = 8:
face 0
61765993
81922273
24373339
57189499
19171363
58549999
94265131
79399393
face 1
81497431
16846339
97367581
33687919
32567137
96261553
32617411
77991313
face 2
66818293
93437353
97742581
76364179
41497177
89431681
55455487
79999333
 

n = 9:


face 0
268768177
684645197
877137563
736152679
972652763
594618457
496779961
591564583
999111917
face 1
523572947
291767947
393118631
958548433
468968447
758197421
343112141
793594957
191933933
face 2
143214287
498521917
656716331
763915687
867781139
394867873
588512531
981242149
737179973

 

n = 10:
 
face 0
4768191851
1215179297
7288419521
3542683667
9335319569
5454869891
1336949879
3279244289
7195498649
1977371993
face 1
1211334521
6663564863
7756541477
6638175653
4981747781
6557614277
1249354727
9894926927
6231595841
7931931971
face 2
5577294293
6645842297
1467929129
3151699667
3938438489
5979698951
3963721571
7646739473
2981626553
3733719371

 

 

***

Oscar Volpatti:

We can build a decent dice with N = 3 by completing the example submitted by Emmanuel Vantieghem.


Solution 1:


 9 2 9     1 8 1
 1 5 7     9 5 3
 9 1 1     1 9 9
 
 7 6 1     3 4 9
 5 4 1     5 6 9
 1 7 3     9 3 7
 
 1 6 3     9 4 7
 2 7 1     8 3 9
 7 7 3     3 3 7
 
Solution 2:
 
 9 2 9     1 8 1
 1 5 7     9 5 3
 9 1 1     1 9 9
 
 7 5 1     3 5 9
 5 4 1     5 6 9
 7 7 3     3 3 7
 
 7 6 1     3 4 9
 4 6 7     6 4 3
 3 1 3     7 9 7
 
Solution 3:
 
 9 2 9     1 8 1
 1 5 7     9 5 3
 9 1 1     1 9 9
 
 7 5 1     3 5 9
 2 8 1     8 2 9
 7 7 3     3 3 7
 
 7 6 1     3 4 9
 4 6 7     6 4 3
 3 1 3     7 9 7
 
Construction.
There are 143 three-digits primes; 128 of them are zero-less.
For 72 such primes p, the three-digits number 1110-p is prime too; we can divide them into 36 pairs (p,1110-p), with p%10 < 5.
We must choose 18 such pairs to fill the opposite faces of the dice (so, any decent dice with N = 3 will have sum 18*1110 = 19980).


In particular, consider the 12 primes needed to fill last row and last column of each matrix:
all their digits must be in C = {1,3,7,9}, as they are the last digits of other primes greater than 5.
There are only 7 candidate pairs:
(191,919), (911,199), (971,139),
(113,997), (173,937), (313,797), (773,337).
Let us label the opposite faces of the dice as 1 and 6, 2 and 5, 3 and 4.
Use two primes in {191,911,971} for face 1;
use 113 and one more prime in {173,313,773} for face 2;
use the remaining two primes in {173,313,773} for face 3.


Now, about 56% of the dice have been filled, in only nine possible ways (up to reflections).
Curiously, if we further require that the top-left digit of every face is in C too,
the only way to complete faces 1 and 6 is the example submitted by Emmanuel Vantieghem;
the remaining four faces can be completed in three ways, as listed above.

***

Emmanuel wrote:

For  N = 5  I found many solutions.
Here, I give you the three faces that have one point in common and whose total of primes is (in my oppinion) minimal :

2 1 1 2 1
1 1 2 5 1
2 8 4 7 7
1 2 2 4 1
1 1 1 1 9

1 5 1 2 1
2 2 4 4 7
2 3 6 8 9
1 2 5 4 7
1 1 3 1 7

3 1 1 2 3
3 1 2 4 9
2 5 1 1 1
1 5 5 8 3
1 1 7 1 9

 
The opposite faces are found by replacing every digit  d  by  10 - d
(and are left to the reader to save space).

 
For  N = 6  there are many solutions.
So, I strengthened the conditions to emirps.
Then, this was wy nicest solution :

 
7 7 9 1 1 1
3 1 2 3 1 1
3 7 3 6 3 1
1 1 4 3 2 9
1 9 1 7 1 7
1 1 1 3 3 7 

7 9 3 9 3 1
9 2 1 5 6 3
7 5 9 5 5 9
1 1 2 9 1 3
3 5 1 5 2 9
1 3 1 7 9 7

9 7 1 1 7 1
1 1 1 1 2 7
3 8 9 9 1 1
3 8 1 9 1 1
3 1 8 8 1 7
1 3 3 3 1 9
 
In every face, the prime in the ith row is the reverse of the prime in the (7-i)th column

***

On March 31, 2020, Michael Hüerter wrote:

I found the following solution for n = 16:
 
face 0
6214737497483717
7689527958288377
6617419861589729
7531423127271629
8699888515612271
9141434641895693
1237644331824353
7285342885865399
3534895129235423
6422869315948853
9374671217468699
2297181768742691
3959474922141449
4757647428176543
8793896779939781
3731739993137939
 
face 1
4348369873661669
6669917647432847
4594736134273829
2577325838296271
9762347689416677
4292365867824857
1711526956951871
6963744168553643
8638123766373887
5164859396563439
9681412812816749
2993546447337227
2341391149871819
1151696847676979
2468377796121563
9999193113333911
 
face 2
6996139693569443
5325878776333463
1264726749454823
2869847833235543
2917943639847749
8556217723388297
6632323699647977
3257285427538271
8763778229622413
3644373318738323
1662143377511291
7473258533252339
5442993593711483
9188826128885219
8746451158296641
3713379999791333

**

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