20061101, 01:55  #78  
Feb 2006
Denmark
230_{10} Posts 
Quote:
If a number n ends in digit d when written in base b, then the number has the form n = k*b+d. If a prime p divides both b and d, then p also divides n. This means the only chance of multidigit primes is when b and d are relatively prime. Dirichlet's theorem says there are infinitely many primes in all such cases. It has also been proved that each d which is relatively prime to n produces the same number of primes asymptotically. 

20061101, 11:59  #79  
May 2006
29_{10} Posts 
Quote:
Thanks for your replies to other mathematicians and to me. I have used +1 as the centre for all primes and prime products, and it has a number of advantages. The expression ((6*M)+1) comprises all primes and prime products, M being any or all of the natural numbers from  infinity to + infinity. ((6*(39)+1) = 233, which´is a prime ((6*(+39)+1) = 235, which is a prime product. A prime product such as ((6*M)+1) * ((6*N)+1) = 36 (NM) + 6*(M+N) + 1 is an integer, which will never be divisible by 2 or 3. Conclusion: 2 and 3 could be called anything but "primes". N (just as M) being any or all natural numbers from  infinity to + infinity. (+) * (+) is of course (+), () * () will also give a (+) integer, (+) * () will give a () integer. All primes and prime products are divisible by 1 (i.e. N=0). If you want to look for (M) and (N), which means to factorize a possible prime, you can do it by subtracting 1 from the integer in question and then use a second order equation to find or not find the two roots (M) and (N). If the sum of (M+N) is odd and > 1, the factorization results in (Even integer)^2  3^2 * (an odd integer)^2. If the sum of (M+N) is 0 or any other even number, the factorization ends in (Odd integer)^2  6^2 * (any integer, including 0)^2. The sign of a prime or prime product can easily be predicted by modulation (modulo 9), and it is easy to show if the sum (M+N) is even or odd. If you have the time you can try to follow my ideas: 7*13 = 91 = 10^23^2 etc. 7*19 = 133 = 13^26^2 etc. I am not drowning. I will in fact consider to reflect to the many harsh replies, which I have received (directly or indirerctly), but maybe it is not worth the effort. A famous citation from Schiller's Jeanne d'Arc comes to my mind (""). Y.s. troels 

20061101, 15:26  #80  
Nov 2003
2^{2}·5·373 Posts 
Quote:
I do not understandand how people can be so totally clueless as to spew the kind of nonsense that has been spewed by this troll. The sad part is that he isn't even aware of how totally clueless his posts have been. 

20061101, 15:38  #81  
Jan 2006
JHB, South Africa
235_{8} Posts 
Quote:
Regards Patrick 

20061101, 21:10  #82 
"Jason Goatcher"
Mar 2005
3×7×167 Posts 
Just as the song goes,"Everybody plays the fool...No exception to the rule..."
EVERYBODY looks stupid at some point, whether they're simply wrong or misunderstood. I'm not attempting a threat in any way, but I would advise people to not have an overly large amount of fun at Mr. Munkner's expense. As someone who tries to stay in tune with the Holy Spirit, I know that sometimes this stuff can pop up again and give us an unpleasant view of ourselves, something we would rather not aknowledge about ourselves. As I reread the above, I realize what I said might not even make sense to Christians, so I'll rephrase it: Sometimes when we judge something unfairly, we can suffer for it later on. 
20061103, 04:11  #83  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Kindly clarify.
Quote:
What most have been confused about is what are the factors of 6.? We know in real numbers that these are 2 x 3. If you dont include these as prime factors according to your definition, as 2 is an even number and 3 is divisible by 3, and these you call 'never primes', then what would you just call them ? [QUOTE =Troels} I am not drowning. I will in fact consider to reflect to the many harsh replies, which I have received (directly or indirerctly), but maybe it is not worth the effort. A famous citation from Schiller's Jeanne d'Arc comes to my mind ("").[/QUOTE] You have mistaken my sentence which I clarified to Mike also. I have not meant that YOU are a drowning man but I can only offer you as much as a drowning man would feel if I threw him a straw. In simple words ' I cannot offer much help in your theory'. I will however endeavour to bring out what you mean and put it more coherently for others to understand Regards Mally 

20061104, 07:25  #84 
May 2006
29 Posts 
Dear Malcolm,
Thanks for your clarification. The expression ((6*M)+1) comprises all primes and prime products, M being any natural number from  infinity to + infinity. But these integers will never be divisible by 2 or 3. Rather soon you will see some new (and important) threads from me. I will appreciate your comments. Y.s. troels 
20061104, 07:55  #85  
May 2006
29 Posts 
Quote:
Whoever you are I can predict that you (sooner or later) will regret some of the replies, which you (directly or indirectly) have sent to me. From your comments I realize that you have limited knowledge of Latin and have not looked into a dictionary with translations of foreign phrases. If you prefer modern, more technical expressions in Esperanto or alike, the translation of "lapsus calami" will be a "typo". It is not worthwhile to react to your other replies. Try to read my threads or replies, openminded for new ideas. When you pretty soon will see new threads on "Fermat's small theorem", "An analysis of (the very few) Mersenne primes and the vast majority of [2^p 1] products", a new tool "SCET (an acronym for Smallest Common Exponential Term, radix 2)" and "Riemann's zetafunction" and maybe want to open these threads, please swallow a couple of tranquillizers before you make your comments. Perhaps you can find a translation (in your dictionary) of the following quotation: "Quousque tandem abutere patientia". Y.s. troels munkner 

20061104, 08:02  #86  
May 2006
29 Posts 
Quote:
Whoever you are I can predict that you (sooner or later) will regret some of the replies, which you (directly or indirectly) have sent to me. From your comments I realize that you have limited knowledge of Latin and have not looked into a dictionary with translations of foreign phrases. If you prefer modern, more technical expressions in Esperanto or alike, the translation of "lapsus calami" will be a "typo". It is not worthwhile to react to your other replies. Try to read my threads or replies, openminded for new ideas. When you pretty soon will see new threads on "Fermat's small theorem", "An analysis of (the very few) Mersenne primes and the vast majority of [2^p 1] products", a new tool "SCET (an acronym for Smallest Common Exponential Term, radix 2)" and "Riemann's zetafunction" and maybe want to open these threads, please swallow a couple of tranquillizers before you make your comments. Perhaps you can find a translation (in your dictionary) of the following quotation: "Quousque tandem abutere patientia". Y.s. troels munkner 

20061104, 10:27  #87 
Feb 2006
Brasília, Brazil
3×71 Posts 
Temporary grammar nazism
It's interesting that Troels criticizes someone else's Latin knowledge and makes such an elementary mistake in the same post.
The correct sentence would be "quousque tandem abutere patientiam". Interestly, it applies perfectly to you, mr. Munkner; how long are you gonna keep coming here, trying to persuade everyone that 2 and 3 aren't prime and that Euclid was a moron? Bruno 
20061107, 15:35  #88  
Jan 2006
52_{8} Posts 
Quote:
I followed this thread for quite a while now and I always asked myself a very simple question: To which of your three groups do the integer numbers 5, 11, 17, 23, ... belong? I hope you can enlighten me. Last fiddled with by Rde on 20061107 at 15:38 Reason: Typo 

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