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#166 |
Jan 2005
Sydney, Australia
5×67 Posts |
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I'm having a problem with the PRPnetclient.exe
It started with an Intel E8400 Wolfdale CPU based system. Now it has occurred on an Intel Q8200 quad-core system also. Both are running Windows 7 Ultimate 64-bit edition. Both computers were happily running NPLB prpnet on port 2000. Now they throw a message "prpclient.exe has stopped working" and below this heading it says "A problem caused the program to stop working correctly. Windows will close the program and notify you if a solution is available." It doesn't. This occurs when the start batch file is initiated. On the quad only 2 of the cores crash, so it leaves 2 instances (in separate sub-directories as per Lennart's batch file) running. I tried re-installing the client using the install batch file right over the top of the existing files. Didn't help. I'm stuck. Any ideas? |
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#167 | |
"Lennart"
Jun 2007
25·5·7 Posts |
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Run install # core then update # core then start and it should work. This is only a fast way to get it running again. You don't have to save the resultfiles if you don't need them. Lennart |
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#168 |
Jan 2005
Sydney, Australia
5×67 Posts |
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Thanks Lennart, that worked for both machines.
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#169 |
Jan 2005
Sydney, Australia
1010011112 Posts |
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Question: Do all Prime Numbers occur next to a number divisible by 6?
Eg 17 is next to 18. |
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#170 |
"Vincent"
Apr 2010
Over the rainbow
B4416 Posts |
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yep, 1 mod 6 , -1 mod 6 or 5 mod 6.
meaning a prime will be of tthe form (x*6) plus or minus one |
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#171 |
Feb 2012
Prague, Czech Republ
3·67 Posts |
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#172 |
Romulan Interpreter
"name field"
Jun 2011
Thailand
101000001000012 Posts |
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#173 |
Jan 2005
Sydney, Australia
5×67 Posts |
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#174 |
Jan 2006
deep in a while-loop
22·167 Posts |
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just for giggles I tested the first 100,000 primes and this is true for
3 < prime <= 239737 However there is no discernible pattern for whether the mod6 'seed' is 1 above or 1 below. I sent you the worksheet ![]() |
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#175 |
"Gary"
May 2007
Overland Park, KS
101110000100012 Posts |
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There is no need to test. It can be easily proven that all primes > 3 must be either p==(1 mod 6) or p==(5 mod 6). Proof:
p==(0 mod 6) is always divisible by 2 and 3 p==(2 mod 6) is always divisible by 2 p==(3 mod 6) is always divisible by 3 p==(4 mod 6) is always divisible by 2 This leaves p==(1 mod 6) and p==(5 mod 6) as the only possible primes mod 6. |
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#176 |
"Daniel Jackson"
May 2011
14285714285714285714
3·251 Posts |
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If you convert the primes into base-6, you'll notice that, except for 2 and 3, they all end in 1 or 5. That's easy proof.
Last fiddled with by Stargate38 on 2012-11-21 at 00:53 |
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