Problems & Puzzles: Puzzles

Puzzle 700 p & q primes such that ...

Emmanuel Vantieghem sent the following nice puzzle

Find pairs or non-palindromic primes (p & q) such that all of the following 12 integers are 12 distinct primes:

p, q, R(p), R(q), p&q, q&p, p&R(q), R(q)&p, R(p)&q, q&R(p),
R(p)&R(q), R(q)&R(p)

One example is this one: p=155599, q=752197.

 p 155599 q 752197 R(p) 995551 R(q) 791257 p&q 155599752197 q&p 752197155599 p&R(q) 155599791257 R(q)&p 791257155599 q&R(p) 752197995551 R(p)&q 995551752197 R(p)&R(q) 995551791257 R(q)&R(p) 791257995551

Emmanuel sent three more examples.

Q. Find the first 10 examples (ordered by p&q).

Contributions came from Giovanni Resta, Jud McCranie, Emmanuel Vantieghem, W. Edwin Clark, Jahngeer Kholdi & Farideh Firoozbakht and Vicente Felipe Izquierdo

***

Giovanni wrote:

Clearly if (p,q) is a solution, also (p,rev(q)),  (rev(p),q), (rev(p),rev(q)) and the 4 other pairs obtained swapping p and q are solutions, since all lead to the same set of 12 numbers.

So I confined the search to pairs for which p<q, p<rev(p) and q<rev(q).

For p,q < 10^7 there are 40 such pairs.
The 10 smallest ones, in terms of  the largest component are:

186007 302647
162391 700459
379909 705247
155599 752197
324143 1089359
12119 1324619
365479 1513591
32299 1548181
72689 1568519
1150927 1589689

I found the smallest pair of consecutive primes
which produce a set of 12 primes. They are:
(1724280032963, 1724280033023)

If one of the two consecutive primes is allowed to be
palindromic (thus reducing the resulting set to 7 primes), then the smallest solution is (70280108207, 70280108219)

***

Jud wrote:

Attached are 50 solutions to puzzle 700.  When listing a reversible prime, I list the smaller number.  The lines are p, q pairs, with q>p.  The list is sorted in order of increasing q.

186007 302647
162391 700459
379909 705247
155599 752197
324143 1089359
12119 1324619
365479 1513591
32299 1548181
72689 1568519
1150927 1589689
74897 1632209
11057 1699307
32939 1845017
13553 1852013
1025209 1862317
1587503 1866593
1311853 1873357
1716499 1879069
197647 1913467
1509887 3023747
16699 3248347
1996229 3289283
301459 3295657
3081523 3319609
1390507 3403339
14923 3513427
1207387 3547909
162017 3637367
336079 3806119
313517 3830429
15383 3841787
3276073 3844747
1353019 3923287
3823649 7020509
31259 7141919
30319 7241029
3117053 7382399
1294369 7576669
7006187 9163079
1860569 9280199
340909 10009603
1004981 10037309
79379 10064081
1963369 10079053
1415207 10209491
7665439 10321639
1576703 10396091
1821649 10548883
3305833 10714831
1266731 10854071

***

Emmanuel wrote:

I computed the first 53 sets of twelve primes.  Each such a set consits of four emirps, say p, q, r(=R(p)), s(=R(q))   and the eight concatenations
p&q, p&s, q&p, q&r, s&p, r&q, s&r, r&s.  All concatenations have at most twelve digits.
Of the concatenations I retain the smallest one and I write it down with a space instead of the symbol '&'.

149 37805387
1031 34240589
1069 11025337
1091 33734849
11057 1699307
113 170699843
115949633 179
12119 1324619
1212722131 13
1223 38711249
1300969741 13
1302807491 17
13 1371020383
13 1613060191
13258151 3023
13 3057444307
1348937531 17
13553 1852013
136562731 337
13 7420441099
1397512849 79
1471 19007017
1475759227 37
14923 3513427
15383 3841787
1548181 32299
155063791 709
155599 752197
1568519 72689
162391 700459
1632209 74897
16699 3248347
16863563 3527
17 7433830367
1845017 32939
186007 302647
1906713169 37
191480039 389
19852747 9349
30319 7241029
308544617 347
3110744863 79
31259 7141919
3140189179 37
3198372427 37
3469023367 79
347 386363249
347 703185257
3526253167 37
359 925232549
3719 38487629
37 9144955489
379909 705247

This list should be 'complete' in the sense that, if someone finds a concatenation (with twelve digits or less) not in this list, then either the concatenation was not the smallest in the corresponding set of twelve primes, or the set does not contain twelve different primes (a mistake that I made first).

Of course, Ii is possible that I made mistakes.  Therefore it would be interesting to see this results doublechecked.

Besides, I conjecture that for every emirp  p  there exist an emirp  q  that will allow the construction (as above) of 12 different emirps.  There is a heuristic argument for that. Anyhow, I verified the conjecture for  p < 1000.

***

Clark wrote:

First, one notices that if (p,q) is a solution then so are all eight of the
primes pairs: { (p,q),  (p,R(q)), (R(p),q), (R(p),R(q), (q,p), (R(q),p),  (q,R(p)), (R(q),R(p))}.
[One can interpret this as an orbit of the dihedral group of order 8 acting on
the set of pairs.]

In all I found 22 distinct orbits of such pairs. Here are representatives of the 10 orbits
with smallest p&q values that I  found: I  would not be surprised if I missed some.

(1 ) 149,37805387
(2 ) 1069,11025337
(3 ) 11057,1699307
(4 ) 113,170699843
(5 ) 115949633,179
(6 ) 12119,1324619
(7 ) 1223,38711249
(8 ) 13258151,3023
(9 ) 13553,1852013
(10) 136562731,337

***

Jahngeer Kholdi & Farideh Firoozbakht wrote:

It is clear if pair (p, q)  is a solution, then all 11 other pairs   (p, R(q)),
(q, p), ... , (R(p), R(q)) have the same property.

The solutions are as follows:

a1.   p = 1031 & q = 34240589
a2.   p = 1031 & q = 98504243
b1.   p = 1069 & q = 11025337
b2.   p = 1069 & q = 73352011
c1.   p = 1091 & q = 33734849
c2.   p = 1091 & q = 94843733
b3.   p = 11025337 & q = 1069
d1.   p = 113 & q = 170699843
d2.   P = 113 & q = 348996071
e1.   p = 12119 & q = 1324619
f1.    p = 1212722131 & q = 13
g1.   p = 1223 & q = 38711249
g2.   p = 1223 & q = 94211783
h1.   p = 1300969741 & q = 13
a3.   p = 1301 & q = 34240589
a4.   p = 1301 & q = 98504243
***
Vicente wrote:

The least primes what I found.

149, 37805387
149, 78350873
941, 37805387
941, 78360873
1031, 34240589
1069, 11025337
1069, 73352011
1091, 33734849
1223, 38711249
1301, 34240589
1471, 19007017
1471, 71070091
1741, 19007017
1741, 77107091
1901, 33734849
3023, 13258151
3023, 15185231

***

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