Problems & Puzzles: Collection 20th

Coll.20th-015. Consecutive primes and Pandigitals

On May 6, 2018, Rodolofo Kurchan, wrote:

Q. Find k consecutive primes that when added produces

a) the smallest ten digits pandigital.

b) the largest ten digits pandigital.

Do this for  k=2, 3, 4,..., 10.

Contributions came from Jeff Heleen, Claudio Meller and Emmanuel Vantieghem

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Jeff wrote on Set 3, 2018:

For Coll.20th-015 I have
Q1:
k = 2: 511729877 + 511729891 = 1023459768
k = 3: 341152541 + 341152571 + 341152577 = 1023457689
k = 4: 255864403 + 255864407 + 255864437 + 255864451 = 1023457698
k = 5: 204697291 + 204697303 + 204697327 + 204697333 + 204697343 = 1023486597
k = 6: 170579861 + 170579897 + 170579939 + 170579951 + 170579957 + 170579963 = 1023479568
k = 7: 146208341 + 146208353 + 146208367 + 146208371 + 146208407 + 146208421 + 146208437 = 1023458697
k = 8: 127933427 + 127933433 + 127933453 + 127933499 + 127933501 + 127933513 + 127933549 + 127933583 = 1023467958
k = 9: 113738491 + 113738507 + 113738531 + 113738621 + 113738641 + 113738671 + 113738683 + 113738701 + 113738743 = 1023647589
k = 10: 102357823 + 102357847 + 102357859 + 102357863 + 102357881 + 102357887 + 102357919 + 102357953 + 102357961 + 102357971 = 1023578964
Q2:
k = 2: 4938271579 + 4938271631 = 9876543210
k = 3: 3292174651 + 3292174687 + 3292174693 = 9876524031
k = 4: 2469112747 + 2469112817 + 2469112829 + 2469112837 = 9876451230
k = 5: 1975300447 + 1975300451 + 1975300493 + 1975300511 + 1975300529 = 9876502431
k = 6: 1646088649 + 1646088673 + 1646088677 + 1646088683 + 1646088707 + 1646088751 = 9876532140
k = 7: 1410934643 + 1410934667 + 1410934687 + 1410934717 + 1410934739 + 1410934781 + 1410934787 = 9876543021
k = 8: 1234542527 + 1234542553 + 1234542607 + 1234542623 + 1234542637 + 1234542679 + 1234542707 + 1234542719 = 9876341052
k = 9: 1097393513 + 1097393519 + 1097393527 + 1097393543 + 1097393551 + 1097393581 + 1097393617 + 1097393623 + 1097393629 = 9876542103
k = 10: 987650239 + 987650263 + 987650311 + 987650317 + 987650341 + 987650369 + 987650371 + 987650387 + 987650401 + 987650413 = 9876503412

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Claudio wrote on Set 3, 2018:

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Emmanuel wrote on 7-9-18:

It was easy to find solutions for  k = 2 to 10 :

Lowest :
k     first prime        first pandigital
2     511729877       1023459768
3     341152541       1023457689
4     255864403       1023457698
5     204697291       1023486597
6     170579861       1023479568
7     146208341       1023458697
8     127933427      1023467958
9     113738491      1023647589
10    102357823      1023578964

Highest
k      first prime         last pandigital
2      4938271579      9876543210
3      3292174651      9876524031
4      2469112747      9876451230
5      1975300447      9876502431
6      1646088649      9876532140
7      1410934643      9876543021
8      1234542527      9876341052
9      1097393513      9876542103
10     987650239        9876503412

It was even relatively easy to do this for all  k <= 26365.
For  k = 26366  there is no solution.  I. e. : no sum of 26366 consecutive primes is (a 10-digit) pandigital.
For  k = 26365 :

Lowest : First prime : 353.  Sum : 3842075961
Highest : First prime : 143821.  Sum : 8104596723

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