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2005-11-10, 16:04
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#12 | |
Jun 2003
13DA16 Posts |
Quote:
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2005-11-10, 18:12
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#13 |
Jun 2003
2·7·113 Posts |
p=2,23,53,853
Are the only case for which sum of first p primes =0 (mod p). (under 5 Million) I have searched the first 5 million primes and am continuing. No solution yet. Citrix edit: Are there any more of this type. The probability is 1/p. (which is not so bad) Last fiddled with by Citrix on 2005-11-10 at 18:27 |
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2005-11-10, 18:39
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#14 |
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Jun 2003
2·7·113 Posts |
checked the first 12.5 Million primes ie primes under 230 Million.
Citrix |
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2005-11-10, 20:10
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#15 |
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Jun 2003
2×7×113 Posts |
The above problem can be solved if some one can prove this or disprove the following. It might already have been conjectured by some one or proved by some one.
For every even number n >4 there exists a prime t < n/2 such that n+t is also prime! (checked to 1M) If this is the case then for us, consider a n=2*p Then t=n+t (mod p) And there are no more solutions to the problem, other than 2. Last fiddled with by Citrix on 2005-11-10 at 20:24 |
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2005-11-10, 20:25
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#16 |
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Jun 2003
110001011102 Posts |
n>4 should be n>7
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2005-11-10, 21:00
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#17 | |
Feb 2005
22·32·7 Posts |
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2005-11-10, 21:36
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#18 |
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Jun 2003
30568 Posts |
Submitted the sequence.
For product ((p(2)#/2)+1)%2 is Zero (0) ((p(241)#/241)+1)%241 is Zero (0) ((p(271)#/271)+1)%271 is Zero (0) ((p(8627)#/8627)+1)%8627 is Zero (0) No more till 200,000. Citrix |
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