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2020-02-18, 12:59
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#1 |
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Mar 2018
2×5×53 Posts |
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 Last fiddled with by enzocreti on 2020-02-18 at 13:34 |
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2020-02-18, 15:44
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#2 |
Feb 2017
Nowhere
4,643 Posts |
You didn't look very hard. Here is a table of consecutive primes p and q < 5000 for which the smallest 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 some exponent k will exist for about a quarter of the primes. k....p...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 Last fiddled with by Dr Sardonicus on 2020-02-18 at 15:48 Reason: Put in bound used for primes |
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2020-02-18, 15:59
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#3 |
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Mar 2018
2·5·53 Posts |
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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? Last fiddled with by enzocreti on 2020-02-18 at 17:00 |
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2020-02-18, 19:10
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#4 | |
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Feb 2017
Nowhere
464310 Posts |
Quote:
39929 39937 43331 43391 66161 66169 
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2020-02-19, 00:28
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#5 | |
Aug 2006
3×1,993 Posts |
Quote:
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2020-02-19, 14:52
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#6 |
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Feb 2017
Nowhere
4,643 Posts |
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 prime 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 |
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