10^k is congruent to p(n)*p(n+1) mod 1093
 

I am searching for consecutive primes p(n) and p(n+1) (p(n) is the n-th prime)


such that




10^k is congruent to p(n)*p(n+1) mod 1093 for some k


for example 10^6 is congruent to 433*439 mod 1093




433
11
103
823
503
479
...




42 primes up to 1000 have this property




10^7 is congruent to (11*13) mod 1093


7 11 and 13 are primes


I have not yet found another k prime such that


10^k is congruent to p(n)*p(n+1) mod 1093

You didn't look very hard. Here is a table of consecutive primes p and q < 5000 for which the [i]smallest[/i] k (k <= 273) such that 10^k == p*q (mod 1093) is prime.

I note that there are two cases in which the exponent k is 2.

Of course, if the smallest exponent k is relatively prime to 273, you can add a multiple of 273 to k to obtain a prime exponent that satisfies the congruence.

My default guess is that [i]some[/i] exponent k will exist for about a quarter of the primes.

k[color=white]....[/color]p[color=white]...[/color]q
149 7 11
7 11 13
31 107 109
233 601 607
103 853 857
229 1031 1033
223 1171 1181
61 1201 1213
193 1667 1669
47 1787 1789
107 2141 2143
19 2239 2243
5 2297 2309
67 2437 2441
157 2521 2531
79 3041 3049
149 3137 3163
71 3191 3203
37 3217 3221
151 3331 3343
149 3733 3739
163 3793 3797
2 3931 3943
127 3947 3967
43 4229 4231
151 4243 4253
79 4591 4597
2 4801 4813
97 4817 4831

...
 

k is 2 in two cases why? just coincidence i think...in the case k=2


the distance between the two consecutive primes=12


That I note

In the case k=2

The primes p are such that p+12 and p-12 are both prime. Maybe just a coincidence

A challenge could be to find a counter example?

[QUOTE=enzocreti;537848]k is 2 in two cases why? just coincidence i think...in the case k=2


the distance between the two consecutive primes=12


That I note

In the case k=2

The primes p are such that p+12 and p-12 are both prime. Maybe just a coincidence

A challenge could be to find a counter example?[/QUOTE]
26813 26821
39929 39937
43331 43391
66161 66169

:sleep:

[QUOTE=Dr Sardonicus;537846]My default guess is that [i]some[/i] exponent k will exist for about a quarter of the primes.[/QUOTE]

It should, yes.

Just for practice, I created a list of prime exponents k for which 10^k == p*q (mod 1093), where p and q are consecutive primes less than 1000. In this list, k can be greater than 273, but is the smallest [i]prime[/i] exponent that works for the given p and q,

k p q
149 7 11
7 11 13
661 23 29
1019 37 41
499 97 101
31 107 109
1627 257 263
1063 353 359
353 463 467
571 479 487
487 499 503
443 571 577
233 601 607
457 641 643
463 751 757
337 809 811
631 811 821
103 853 857
941 983 991