Problems & Puzzles: Puzzles

Puzzle 1051. About consecutive integers whose reverse are primes. Paolo Lava sent the following nice puzzle. Let us consider two consecutive integers whose reverse are primes. E. g. 13 and 14 because 31 and 41 are both primes The first pairs are (2,3), (13,14), (16,17), (31,32), (34,35), (37,38), (70,71), (74,74), (91,92), (97,98), (106,107), (112,113), (118,119), etc. I was looking to records of consecutive pairs (jumping over these n@3=0 values). To be clear, (13,14), (16,17) are 2 consecutive pairs while (31,32), (34,35), (37,38) are 3 consecutive pairs.   Here below the first 4 consecutive pairs.   Number of consecutive pairs Pairs 4 (1099, 1100), (1102, 1103), (1105, 1106), (1108, 1109) 4 (72598, 72599), (72601, 72602), (72604, 72605), (72607, 72608) 4 (7473208, 7473209), (7473211, 7473212), (7473214, 7473215), (7473217, 7473218) 4 (16226557, 16226558), (16226560, 16226561), (16226563, 16226564), (16226566, 16226567) 4 (16226557, 16226558), (16226560, 16226561), (16226563, 16226564), (16226566, 16226567) 4 (30528955, 30528956), (30528958, 30528959), (30528961, 30528962), (30528964, 30528965) 4 (73057834, 73057835), (73057837, 73057838), (73057840, 73057841), (73057843, 73057844) 4 (78146635, 78146636), (78146638, 78146639), (78146641, 78146642), (78146644, 78146645) 4 (79015591, 79015592), (79015594, 79015595), (79015597,79015598), (79015600, 79015601) 4 (93519021, 93519023), (93519025, 93519026), (93519028, 93519029), (93519031, 93519032) 4 (96448798,96448799), (96448801,96448802), (96448804,96448805), (96448807, 96448808)   Here below the first 5 consecutive pairs.   Number of consecutive pairs Pairs 5 (751, 752), (754, 755), (757, 758), (760, 761), (763, 764) 5 (17153065, 17153066), (17153068, 17153069), (17153071, 17153072), (17153074, 17153075), (17153077, 17153078) 5 (131127805, 131127806), (131127808, 131127809), (131127811, 131127812), (131127814, 131127815), (131127817, 131127818)   Q. Are there number of  consecutive pairs greater then 5?

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Contributions during the week 22-27 August 2021, came from Oscar Volpatti, Emmanuel Vantieghem, Giorgos Kalogeropoulos, Simon Cavegn

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Oscar, Emmanuel, Giorgios and Simon found the same and first set of six consecutive pairs:

(97971007093,97971007094)

(97971007096,97971007097)

(97971007099,97971007100)

(97971007102,97971007103)   

(97971007105,97971007106)

(97971007108,97971007109)

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Oscar added:

Claim 2:  there are no solutions with more than six consecutive pairs.

Consider a number n divisible by 11 but not by 3, so that n belongs to exactly one pair; we can distinguish two cases.

n == 1 mod 3, so n = 33*q+22; n belongs to the type-1 pair (33*q+22, 33*q+23).
n == 2 mod 3, so n = 33*q+11; n belongs to the type-2 pair (33*q+10, 33*q+11).

Let's recall the divisibility-by-11 rule in base 10: sum of even-position digits == sum of odd-position digits (mod 11).

Therefore 11 divides reverse(n) too, so that reverse(n) is composite, and the pair containing n must be discarded, unless we have reverse(n) = 11.
Equality holds for infinitely many numbers of the special form n = 11*10^k:  11, 110, 1100, ...

Congruence  11*10^k == 1*2+0*k == 2 (mod 3)  holds, so k-th special number belongs to a type-2 pair, with index q = (10^k - 1)/3:

for k = 0, we get q = 0;
for k > 0, we get q = 3*R(k), a 3-repdigit with length k.
Therefore we must discard all type-1 pairs, along with all type-2 pairs whose index is neither 0 nor a 3-repdigit.

What about the remaining type-2 pairs, of the special form (11*10^k - 1, 11*10^k) ?

We must discard most of them too, because reverse(11*10^k - 1) is prime or PRP only for k = 2, 3, 5, 25, 89, 246, 345, 499, 533, 765, 2000, 2579, 2675, 8156, ...

 

After discarding all type-1 pairs, we can have only blocks of at most ten consecutive pairs; q-th block has the form:
 

(33*q-8, 33*q-7) ... (33*q+19, 33*q+20).

If the type-2 pair (33*q+10, 33*q+11) is discarded, then q-th block actually contains at most six consecutive pairs:

(33*q-8, 33*q-7) ... (33*q+7, 33*q+8). 

By the way, my smallest solution of length six has this form, with index q = 2968818397.

On the other hand, all blocks whose type-2 pair is not discarded are covered by the following stronger bound:
 

if index q is any 3-repdigit of length k >= 2, then q-th block contains at most five consecutive pairs.
 

 

PROOF.

For k >= 2, the twenty reverses admit the following simple representation as a function of k:
 

(19*10^k - 99, 29*10^k - 99),

(49*10^k - 99, 59*10^k - 99),

(79*10^k - 99, 89*10^k - 99),

(10*10^k - 99, 20*10^k - 99),

(40*10^k - 99, 50*10^k - 99),

(70*10^k - 99, 80*10^k - 99),

(100*10^k - 99, 0*10^k + 11),

(20*10^k + 11, 30*10^k + 11),

(50*10^k + 11, 60*10^k + 11),

(80*10^k + 11, 90*10^k + 11).

Let's check divisibility of reverses by 7 and by 13.

As 10^6 == 1 mod 7*13, both divisibility patterns repeat with period 6, so we only need to check from k = 2 to k = 7.

I'll mark reverses in red if divisible by 7, in green if divisible by 13, in blu if divisible by 7*13.

 

k = 2, at most five consecutive pairs remaining.
 

(1801, 2801),

(4801, 5801),

(7801, 8801),

(901, 1901),

(3901, 4901),

(6901, 7901),

(9901, 11),

(2011, 3011),

(5011, 6011),

(8011, 9011).

 

k = 3, at most three consecutive pairs remaining.

(18901, 28901),

(48901, 58901),

(78901, 88901),

(9901, 19901),

(39901, 49901),

(69901, 79901),

(99901, 11),

(20011, 30011),

(50011, 60011),

(80011, 90011).

 

k = 4, at most two consecutive pairs remaining.

(189901, 289901),

(489901, 589901),

(789901, 889901),

(99901, 199901),

(399901, 499901),

(699901, 799901),

(999901, 11),

(200011, 300011),

(500011, 600011),

(800011, 900011).

 

k = 5, at most two consecutive pairs remaining.

(1899901, 2899901),

(4899901, 5899901),

(7899901, 8899901),

(999901, 1999901),

(3999901, 4999901),

(6999901, 7999901),

(9999901, 11),

(2000011, 3000011),

(5000011, 6000011),

(8000011, 9000011).

 

k = 6, at most four consecutive pairs remaining.

(18999901, 28999901),

(48999901, 58999901),

(78999901, 88999901),

(9999901, 19999901),

(39999901, 49999901),

(69999901, 79999901),

(99999901, 11),

(20000011, 30000011),

(50000011, 60000011),

(80000011, 90000011).

 

k = 7, at most three consecutive pairs remaining.

(189999901, 289999901),

(489999901, 589999901),

(789999901, 889999901),

(99999901, 199999901),

(399999901, 499999901),

(699999901, 799999901),

(999999901, 11),

(200000011, 300000011),

(500000011, 600000011),

(800000011, 900000011).

 

So we can have at most five consecutive pairs, when k is congruent to 2 mod 6:

(11*10^k - 4, 11*10^k - 3)...(11*10^k + 8, 11*10^k + 9).

QED.

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Emmanuel added:

It is not possible to find seven such duo's.
For it is easy to show that in the set {m, m+1, m+3, m+4, m+6, m+7, m+9, m+10, m+12, m+13, m+15, m+16, m+18, m+19} there is always at least one element divisible by three of eleven.
The reverse of such an element is then also divisible by three or eleven.

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