20110829, 17:07  #1 
May 2011
France
7·23 Posts 
Power???
I need to find the first multiple of a prime number > an offset
i/e Let Offset be 223092870 Let P be 21247 I want N like Offfset+N = 223 093 500 = 105000 * 12247 223092870 div 21247 = 10499 223093500 223092870 = 630 = N I code this but I have a problem when the offset change I do N:=P(offset mod p) Is the problem in the pollynome ? Thanks 
20110829, 18:53  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
2)the underline is a line I personally don't see the use for 3)the italic is a spelling error 4) really because the line I wrapped in code tags says (Offset+N)Offset = N 5) I come to: X=Offset+(P(Offset%P)) working no ? 

20110830, 06:33  #3 
May 2011
France
7·23 Posts 
Typo
( yes I've checked it, and I'm guessing it's a typo?)
Yes I inverse the two first digir of P 21247 is not 12247!!!! 223092870 div 21247 = 10499 P^10499 < Offset p^10500 > Offset N= P^10500 Offset or N=P(offset mod p) N must be >0 and < Offset) I restart from 0 and I code verifying the error line by line Have you an idea how to process the Golback conjecture: all even value is the sum of two primes numbers?? John 
20110830, 11:29  #4  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
Code:
f=[];for(i=1,100,for(j=1,100,f=concat(f,if((prime(i)+prime(j))%2==0,prime(i)+prime(j)))));vecsort(f,,8) 

20110830, 15:55  #5  
May 2011
France
7×23 Posts 
Find!
The polynome was goof the error was at the line just after.
I make N= N^2 ((an optimization) but it's good only if you begin at one: Quote:
You can modify 3 Starting from p^2, count up in increments of p and mark each of these numbers greater than p itself in the list. These numbers will be 2p, 3p, 4p, etc.; note that some of them may have already been marked. not if you make a continue search not a set of continues integers John 

20110830, 16:28  #6  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
if you are looking to make a code I'll give you the steps but you likely don't want that. 

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