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Digit sums 9 messages
Hi All Here is some interesting things to examine. We all know that if the digit sum of a number is 9 then the number is divisible by 9. So 0% of these numbers are prime But what if the digit sum is 8
Hi All Here is some interesting things to examine. We all know that if the digit sum of a number is 9 then the number is divisible by 9. So 0% of these numbers are prime But what if the digit sum is 8
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By kimowww
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RFDS prize quiz contest 2 messages
If you’re reading this, I really hope you will enter my new high-IQ-like quiz contest. For two reasons: first, there is a significant cash prize (US$100) for the winner; and second, and more important
If you’re reading this, I really hope you will enter my new high-IQ-like quiz contest. For two reasons: first, there is a significant cash prize (US$100) for the winner; and second, and more important
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By Tim Roberts
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Probability that n random walks (1D) intersect a single point 16 messages
Hello, Last weeks I did an analysis on the intersection of n-random walks in a single point. There are quit some new (to me) observations I made. A short question is posted on mathematics stacks excha
Hello, Last weeks I did an analysis on the intersection of n-random walks in a single point. There are quit some new (to me) observations I made. A short question is posted on mathematics stacks excha
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By OOOVincentOOO
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Formula for number of coprimes in 2N interval
Hi All I wonder if there exist a formula that can calculate the number of coprimes in a 2N interval by using the positions of the primes. One could use a empirical approach to this problem. Suppose th
Hi All I wonder if there exist a formula that can calculate the number of coprimes in a 2N interval by using the positions of the primes. One could use a empirical approach to this problem. Suppose th
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By kimowww
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Factoring large difficult-to-factor positive integers
Thank you, Tim. I see from the smallest three numbers that you posted that none of my own constructed factoring algorithms can compete with Pollard Rho for factoring the large difficult-to-factor posi
Thank you, Tim. I see from the smallest three numbers that you posted that none of my own constructed factoring algorithms can compete with Pollard Rho for factoring the large difficult-to-factor posi
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By Kermit Rose
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factorization 2 messages
Hi all, I've been asked by a correspondent for some large numbers that can be used to test their factorization method. This may be useful to others in this group, so here are a few I've concocted: 723
Hi all, I've been asked by a correspondent for some large numbers that can be used to test their factorization method. This may be useful to others in this group, so here are a few I've concocted: 723
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By Tim Roberts
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THE ORANGE THEORY - Mathematical Calculations of Odds and Index of Coincidences needed
This is a report on the constant patterning of use of a specific ciphertext strand and template that holds for the Z340 and other ciphers in the mix Included is the use of the steganography of Pi in t
This is a report on the constant patterning of use of a specific ciphertext strand and template that holds for the Z340 and other ciphers in the mix Included is the use of the steganography of Pi in t
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By Eldorado
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Difficulty of factoring positive integers
Currently, I expect the difficulty, in general, of factoring an odd positive integer to be proportional to the cube of its logarithm. Kermit
Currently, I expect the difficulty, in general, of factoring an odd positive integer to be proportional to the cube of its logarithm. Kermit
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By Kermit Rose
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Circle from n-gon circumference=1
Hello, Usually when determining pi a circle is drawn with a radius, then polygons are inscribed and out scribed. Lasts months I attempted to find a more intuitive way explaining the irrational and tra
Hello, Usually when determining pi a circle is drawn with a radius, then polygons are inscribed and out scribed. Lasts months I attempted to find a more intuitive way explaining the irrational and tra
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By OOOVincentOOO
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Something interesting 2 messages
Hi All Here is something, I find interesting. It looks like the following is true with some exceptions : Let N^2<N^2+r<(N+1)^2 then (N+k)^2+r+k is either prime ( possibly it is only coprime with the p
Hi All Here is something, I find interesting. It looks like the following is true with some exceptions : Let N^2<N^2+r<(N+1)^2 then (N+k)^2+r+k is either prime ( possibly it is only coprime with the p
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By kimowww
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Integer solutions to these equations 5 messages
Hi All I wonder if it is possible to find the integer solutions to diophantine equations of this form : Z=N*(2x+1)-((N+1+2x)/2)*y Example : N=5 Z=5*(2x+1)-((6+2x)/2)*y=10x+5-3y-x*y I want to find all
Hi All I wonder if it is possible to find the integer solutions to diophantine equations of this form : Z=N*(2x+1)-((N+1+2x)/2)*y Example : N=5 Z=5*(2x+1)-((6+2x)/2)*y=10x+5-3y-x*y I want to find all
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By kimowww
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How efficient would this factoring algorithm be for factoring large integers? 5 messages
How efficient would this factoring algorithm be for factoring large integers? z = 33 (z-1) = 32 t1 *(t1 * t2 * t3 + t2 + t3) = 32 t1 = 2 (t1*t2*t3 + t2 + t3) = 16 (2*t2*t3+t2+t3)=16 t2 and t3 are same
How efficient would this factoring algorithm be for factoring large integers? z = 33 (z-1) = 32 t1 *(t1 * t2 * t3 + t2 + t3) = 32 t1 = 2 (t1*t2*t3 + t2 + t3) = 16 (2*t2*t3+t2+t3)=16 t2 and t3 are same
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By Kermit Rose
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An illustration of the complicated relationship between addition and multiplication. 9 messages
An illustration of the complicated relationship between addition and multiplication. 10*30 + 1*10 + 1*30 +1*1 = 341 10*29 +2*10 + 1*29 + 1*2 = 341 9*30 + 1*9 + 2*30 +2*1 = 341 10*28 + 3*10 + 1*28 +1*3
An illustration of the complicated relationship between addition and multiplication. 10*30 + 1*10 + 1*30 +1*1 = 341 10*29 +2*10 + 1*29 + 1*2 = 341 9*30 + 1*9 + 2*30 +2*1 = 341 10*28 + 3*10 + 1*28 +1*3
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By Kermit Rose
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An interesting identity 6 messages
An interesting identity: ((x−a)(x−b)) / ((c−a)(c−b)) + ((x−b)(x−c)) / ((a−b)(a−c)) + ((x−c)(x−a)) / ((b−c)(b−a))=1
An interesting identity: ((x−a)(x−b)) / ((c−a)(c−b)) + ((x−b)(x−c)) / ((a−b)(a−c)) + ((x−c)(x−a)) / ((b−c)(b−a))=1
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By Kermit Rose
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Proposed partial proof of the binary Goldbach conjecture
Dear colleagues, Attached is my manuscript containing a proposed partial proof of the binary Goldbach conjecture. Your constructive comments are most welcome. Best regards, Tatenda Kubalalika.
Dear colleagues, Attached is my manuscript containing a proposed partial proof of the binary Goldbach conjecture. Your constructive comments are most welcome. Best regards, Tatenda Kubalalika.
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By tatendakubalalika@...
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FW: Testing my factoring of integers program
>>> z=(z-1)*53*59+1 >>> Factor2021(z) x = 277 y = 6941373105249865123 z = 1922760350154212639071 Number of iterations = 2 Probable Prime Test Algorithm Factor, using base, 3 >>> z=(z-1)*61*67+1 >>> Fa
>>> z=(z-1)*53*59+1 >>> Factor2021(z) x = 277 y = 6941373105249865123 z = 1922760350154212639071 Number of iterations = 2 Probable Prime Test Algorithm Factor, using base, 3 >>> z=(z-1)*61*67+1 >>> Fa
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By Kermit Rose
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Testing my factoring of integers program
Yesterday, I thought of this "square root algorithm", but the test shows that it is not as good as the older probable prime test factor algorithm. >>> z=2*3*5*7*11*13*17*19+1 >>> Factor2021(z) x = 347
Yesterday, I thought of this "square root algorithm", but the test shows that it is not as good as the older probable prime test factor algorithm. >>> z=2*3*5*7*11*13*17*19+1 >>> Factor2021(z) x = 347
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By Kermit Rose
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Problem in algebra 11 messages
I'm having difficulty expanding the following expression I want to write it as a 4th degree polynomial in y1 so that I can take the next step of assigning functional values to y1 that will make z2^2 a
I'm having difficulty expanding the following expression I want to write it as a 4th degree polynomial in y1 so that I can take the next step of assigning functional values to y1 that will make z2^2 a
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By Kermit Rose
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3x3 magic square of squares solved (without graphics or spheres): 9 messages
take the perfect solution: a|b|c d|e|f g|h|i which yields (f.p.p) : 2e = c + g [0] The perfect solution above can be re-written: b|a|c h|g|i e|d|f which yields 2g = c+e [1] subtracting [0] & [1] toget
take the perfect solution: a|b|c d|e|f g|h|i which yields (f.p.p) : 2e = c + g [0] The perfect solution above can be re-written: b|a|c h|g|i e|d|f which yields 2g = c+e [1] subtracting [0] & [1] toget
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By Conor Williams
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An Inequality related to Legendre conjecture 2 messages
Hi All I hope someone will help me out here. Legendre conjecture is true if this is true (see attachment). This involves a summation series, which is using Ln(N). Kim
Hi All I hope someone will help me out here. Legendre conjecture is true if this is true (see attachment). This involves a summation series, which is using Ln(N). Kim
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By kimowww
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