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20181229, 19:08  #89  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
\[ 1^w7^x17^y23^z\bmod(24)\equiv 1^w7^{\min(y,z)+x}17^{y\min(y,z)}23^{z\min(y,z)}\] and \[\max(y,z)\min(y,z)\equiv 0 \bmod 2\] and \[x+\min(y,z)\equiv 1 \bmod 2\] probably more but I'll leave it for now. Last fiddled with by science_man_88 on 20181229 at 19:10 

20181230, 13:52  #90  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:
\[1^a7^b17^c23^d31^e41^f47^g49^h71^i73^j79^k89^l97^m103^n113^o119^p\] and \[\max(c+f+l+o,d+g+i+p)\min(c+f+l+o,d+g+i+p)\equiv 0\bmod 2\] and \[b+e+k+n +\min(c+f+l+o,d+g+i+p)\equiv 1 \bmod 2 \] As to what else can be said I'm not sure. Last fiddled with by science_man_88 on 20181230 at 13:54 

20181230, 15:16  #91 
"Jeppe"
Jan 2016
Denmark
5·31 Posts 
I am not sure I understand. What are \(x\), \(y\) and \(z\) here? If one of \(y\) and \(z\) is odd and the other one even, the first one is false. And the second one is false if I pick the parity of \(x\) (in)correctly. I may be missing some context? /JeppeSN

20181230, 15:20  #92  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
appear in fully factored mersenne numbers. Last fiddled with by science_man_88 on 20181230 at 15:42 

20181230, 18:32  #93 
Aug 2006
13445_{8} Posts 

20181230, 18:56  #94  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
I was just giving natural powers to the possible remainders of prime factors, and saying that since any factorization even using conposites needs the power on 7 to be odd in the end. since 17 times 23 is 7 mod 24 we can pair them up and get rid of the minimal one used. we then have 17^2 and 23^2 being 1 mod 24 so the difference in powers needs be even. Last fiddled with by science_man_88 on 20181230 at 18:56 

20190107, 07:28  #95  
Aug 2006
3·5^{2}·79 Posts 
Quote:
Code:
atleast(x,lambda)=1sum(k=0,ceil(x)1,lambda^k/k!)/exp(lambda); addhelp(atleast, "atleast(x,lambda): Given a Poisson event with parameter lambda, finds the probability of getting at least x events."); atleast(13, exp(Euler)*log(10)/log(2)) 

20190107, 07:51  #96  
"Mihai Preda"
Apr 2015
1221_{10} Posts 
Quote:
And the same formula (if I'm using it correctly) seems to indicate a 38% probability of finding at least one more between 83M and 100M atleast(1, exp(Euler)*log(100/83.)/log(2)) == 0.38 

20190107, 15:02  #97 
Aug 2006
3×5^{2}×79 Posts 

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