Problems & Puzzles: Puzzles

Puzzle 522. Three sets of k consecutive primes JM Bergot poses the following nice puzzle: Take three sets of four consecutive primes {2, 3, 5, 7}, {17, 19, 23, 29}, and {127, 131, 137, 139} with no two the same, to get 2|(17+127); 3|(19+131), 5|(23+137) and 7|(29+139). Q.  Can you find three sets with more than four consecutive primes, showing the same property?

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Contributions came from A.  Verroken, Torbjörn Alm, Farideh Firoozbakht, Farid Lian & Frederick Schneider:

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Antoine wrote:

Please will find below a solution to p 522 :

2,3,5,7,11 -29,31,37,41,43- 181,191,193,197,199
2,3,5,7,11 -37,41,43,47,53- 149,151,157,163,167
2,3,5,7,11 -41,43,47,53,59- 251,257,263,269,271
3,5,7,11,13 -43,47,53,59,61- 257,263,269,271,277

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Torjjörn wrote:

For starting prime = 2, I have found the following sets size 10:

Size: 10
divisors: 2 3 5 7 11 13 17 19 23 29
Group1: 56453 56467 56473 56477 56479 56489 56501 56503 56509 56519
Group2: 222323 222329 222337 222347 222349 222361 222367 222379 222389 222403
Size: 10
divisors: 2 3 5 7 11 13 17 19 23 29
Group1: 132739 132749 132751 132757 132761 132763 132817 132833 132851 132857
Group2: 289733 289741 289759 289763 289771 289789 289837 289841 289843 289847
Size: 10
divisors: 2 3 5 7 11 13 17 19 23 29
Group1: 214391 214399 214433 214439 214451 214457 214463 214469 214481 214483
Group2: 1162789 1162793 1162807 1162853 1162859 1162867 1162877 1162879 1162897 1162901
Size: 10
divisors: 2 3 5 7 11 13 17 19 23 29
Group1: 785573 785579 785591 785597 785623 785627 785641 785651 785671 785693
Group2: 1218683 1218691 1218709 1218727 1218731 1218739 1218761 1218773 1218779 1218787

...

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Farideh wrote:

The smallest three sets with k consecutive primes, 3*k distinct primes and the
mentioned property for k = 2, 3, ..., 8 that I found are as follows.

k = 2 : {2, 3}, {5, 7}, {13, 17}

k = 3 : {2, 3, 5}, {7, 11, 13}, {29, 31, 37}

k = 4 : {2, 3, 5, 7}, {11, 13, 17, 19}, {67, 71, 73, 79}

k = 5 : {2, 3, 5, 7, 11}, {19, 23, 29, 31, 37}, {113, 127, 131, 137, 139}

k = 6 : {2, 3, 5, 7, 11, 13}, {41, 43, 47, 53, 59, 61}, {251, 257, 263, 269, 271, 277}

k = 7 : {2, 3, 5, 7, 11, 13, 17}, {461, 463, 467, 479, 487, 491, 499},
{3803, 3821, 3823, 3833, 3847, 3851, 3853}

k = 8 : {2, 3, 5, 7, 11, 13, 17, 19}, {3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259},
{23327, 23333, 23339, 23357, 23369, 23371, 23399, 23417}
 

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Farid wrote:

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n=8, examples
2,3,5,7,11,13,17,19
3209,3217,3221,3229,3251,3253,3257,3259
23327,23333,23339,23357,23369,23371,23399,23417
2,3,5,7,11,13,17,19
4409,4421,4423,4441,4447,4451,4457,4463
19139,19141,19157,19163,19181,19183,19207,19211
2,3,5,7,11,13,17,19
14389,14401,14407,14411,14419,14423,14431,14437
56957,56963,56983,56989,56993,56999,57037,57041
2,3,5,7,11,13,17,19
15629,15641,15643,15647,15649,15661,15667,15671
52583,52609,52627,52631,52639,52667,52673,52691
2,3,5,7,11,13,17,19
28979,29009,29017,29021,29023,29027,29033,29059
42929,42937,42943,42953,42961,42967,42979,42989

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Fred wrote:

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Solution for 31 : 1572217 and 4294223
1572217+4294223 = 0 (mod 31)
1572203+4294207 = 0 (mod 29)
1572191+4294189 = 0 (mod 23)
1572187+4294177 = 0 (mod 19)
1572163+4294163 = 0 (mod 17)
1572149+4294153 = 0 (mod 13)
1572113+4294099 = 0 (mod 11)
1572101+4294067 = 0 (mod 7)
1572097+4294051 = 3 (mod 5)
1572091+4294039 = 2 (mod 3)
1572083+4293941 = 0 (mod 2)

Solution for 37 : 2603099 and 66713663
2603099+66713663 = 0 (mod 37)
2603093+66713651 = 0 (mod 31)
2603087+66713641 = 0 (mod 29)
2603077+66713587 = 0 (mod 23)
2603063+66713573 = 0 (mod 19)
2603059+66713557 = 0 (mod 17)
2603047+66713551 = 0 (mod 13)
2603033+66713533 = 0 (mod 11)
2602993+66713527 = 0 (mod 7)
2602973+66713497 = 0 (mod 5)
2602939+66713477 = 0 (mod 3)
2602937+66713473 = 0 (mod 2)

All solutions are sum-wise minimal (except for 37's solution which may not be).

Implemented in Perl (of all things) with a multi-dimensional hash

Later he added:

 converted my lousy Perl script and streamlined it in C++.

Below is the minimal answer for 37. Plus, I found the minimal answers
for 41 and 43. If you could update the page with these solutions,
that'd be great.

Best, Fred

7952389 + 19677731 = 0 (mod 37)
7952369 + 19677683 = 0 (mod 31)
7952363 + 19677677 = 0 (mod 29)
7952333 + 19677659 = 0 (mod 23)
7952309 + 19677643 = 0 (mod 19)
7952267 + 19677629 = 0 (mod 17)
7952249 + 19677613 = 0 (mod 13)
7952233 + 19677611 = 0 (mod 11)
7952183 + 19677601 = 0 (mod 7)
7952171 + 19677589 = 0 (mod 5)
7952129 + 19677583 = 0 (mod 3)
7952123 + 19677571 = 0 (mod 2)

139967879 + 275640511 = 0 (mod 41)
139967837 + 275640509 = 0 (mod 37)
139967827 + 275640493 = 0 (mod 31)
139967819 + 275640461 = 0 (mod 29)
139967801 + 275640451 = 0 (mod 23)
139967771 + 275640433 = 0 (mod 19)
139967761 + 275640317 = 0 (mod 17)
139967743 + 275640307 = 0 (mod 13)
139967701 + 275640283 = 0 (mod 11)
139967689 + 275640259 = 0 (mod 7)
139967669 + 275640241 = 0 (mod 5)
139967657 + 275640229 = 0 (mod 3)
139967651 + 275640221 = 0 (mod 2)

171146077 + 892667989 = 0 (mod 43)
171146057 + 892667987 = 0 (mod 41)
171146051 + 892667959 = 0 (mod 37)
171146047 + 892667933 = 0 (mod 31)
171146039 + 892667921 = 0 (mod 29)
171146029 + 892667911 = 0 (mod 23)
171146011 + 892667903 = 0 (mod 19)
171145999 + 892667863 = 0 (mod 17)
171145993 + 892667849 = 0 (mod 13)
171145991 + 892667837 = 0 (mod 11)
171145981 + 892667833 = 0 (mod 7)
171145973 + 892667777 = 0 (mod 5)
171145963 + 892667759 = 0 (mod 3)
171145951 + 892667701 = 0 (mod 2)

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