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.

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

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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.

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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

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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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On April 10, 2020, Emmanuel Vantieghem added:

I could find a bit more about 'decent dices'.

 

For N = 6, there are more dices with
-  all faces give emirps
-  every face is skew symmetric ( i. e. : symmetric with respect to the 2nd diagonal
-  the second diagonal is also an emirp (you may read it in any direction)
The first diagonal however cannot be prime (since it is palindromic)
Here is one in wihich every face has first diagonal semiprime (11 times a palprime) :

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

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

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

There is no emirp dice with skew symmetric faces when  N = 7.

I found an emirp dice for  N = 8  with skew symmetric faces
(but the second diagonal fails to be prime) :

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

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

face 3 :
1 3 9 7 1 3 3 7
7 1 1 9 6 6 7 3
9 4 3 9 6 6 6 3
3 5 5 3 2 6 6 1
3 6 9 3 3 9 9 7
9 5 5 9 5 3 1 9
1 3 5 6 5 4 1 3
7 1 9 3 3 9 7 1

 

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