.11266 === Subject: JSH: Tired of the deluded posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Unfortunately the truth hurts for some people. I have no doubt that there will be posters who will reply just to clog up threads in the hope that by babbling a lot they can hide the simple solution to the factoring problem And its simplicity should tell you something. As I've said the modern math field is corrupted. Modern mathematician in pure math areas are usually not doing ANY mathematics of value so it's not surprising that they missed something this easy. The insults and accusations of mental illness are just the tools of con artists as no legitimate researcher would need to stoop to such lows. I'm using the factoring problem to prove this reality which has put the future scientific progress of the entire human race in jeopardy. These people are parasitic. They rely on manipulation and attempt to use trust to keep their victims in their control. They do fake math so the answer I have to them as they dominate this earth so strongly is to use real math to show the world what they are, and who they are. Yes, George W. Bush is one of them in the political arena. They are a parasitic subspecies of humanity. They live to manipulate and control without giving value, just like parasites in other arenas. And they have few if any morals. But they are very adept at manipulating people. That's what they live for, it's their greatest talent--mind control. James Harris === Subject: Re: JSH: Tired of the deluded math hurts the one you love....... Simpletons do simplicity. that is what you say, alright, but you are the one that keeps making mistakes. who is they ? got any specifics or are you just babbling? We, the Hive, the Swarm, the Parasites, the Cows continue to keep you, JSH math con artist blocked. We cannot cure you, until you know comples numbers. BS, banks do not use RSA. They do not have to Factor anything. You are stuck back in the 1960's. no! We are the Cows. MOOOOOOOOOOOOOOOOOOOOOOO! We stomple on your feeble and pathetic attempts at math. *****Projection detected:***** I do fake math so the answer you have to me as I dominate this earth so strongly is to use real math to show the world what I am, and who I am. like GWB is going to save your sorry mathass.... *****Projection detected:***** I am a parasitic subspecies of humanity. *****Projection detected:***** I live to manipulate and control without giving value, just like parasites in other arenas *****Projection detected:***** And I have few if any morals. *****Projection detected:***** But I think I am very adept at manipulating people. That's what I live for, it's My greatest talent--mind control *******DUFUS TROLL DETECTED********** === Subject: Re: JSH: Tired of the deluded posting-account=rIfu6QoAAAD5nXG3h9QEE0J3dZn1U45R [...] Please don't crosspost your amusing counter-rants to sci.physics. k? === Subject: Re: JSH: Tired of the deluded no problem, the ROS does not need JSH turd-lets plugging up that pipe, but alt.math.underground does. Solutins Manuals my ass........ === Subject: Re: JSH: Tired of the deluded posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Although many references have been made to the James Harris drinking game, the fact is that the protracted nature of usenet threads makes playing this game impractical for non-homeless posters. I therefore propose instead the game of James Harris bingo. The rules are simple: when James starts a new thread, simply draw up a 5*5 grid containing randomly generated integers from 1 to 50. Then, whenever James uses one of the listed JSH clich.8es, cross off one incidence the corresponding number from your grid, if there is one. When you have crossed off a line of five numbers (horizontal, vertical or diagonal) stand up and shout HOUSE, then attempt to explain to anybody who witnesses this why you did so. The list follows: 1 Brilliant 2 Genius 3 Blocking 4 Lying 5 Inappropriate use of a comma 6 Responding to a counterexample with a modified conjecture which is disproved by the same counterexample 7 'Answering' a yes-or-no question without using the words yes or no 9 My alma-mater Vanderbilt University 10 Introducing new terminology 11 Answering requests to define terminology with an example 12 Answering requests to define terminology with an accusation that the asker is pretending not to know the answer 13 Freaking 14 Beautiful 15 Bizarre 16 Folks 17 Claim that he is performing an experiment on Usenet posters 18 The Hammer 19 THE [sic] prime counting function 20 The object ring 21 Cum Hoc Ergo Propter Hoc 22 American class warfare 23 Fascinating 24 Claim to have solved the factoring problem 25 ...but refuses to factor a number because that would destroy civilization 26 ...but refuses to factor a number because the secret service would kill him 27 ...but refuses to factor a number because the proof is trivial 28 Social crap 29 Social crap 30 Arguing against the distributive property 31 Galois theory as currently taught 32 Double hyphen to indicate a dramatic pause 33 Using the word remind as if it were intransitive 34 White collar welfare 35 Referring to another poster using scare quotes 36 Death threat 37 The math doesn't care 38 sub-human 39 Out to infinity 40 Comparison of self to Gauss 41 Reference to John Nash or Britney Gallivan 42 properly a unit 43 Claim that the ring of algebraic integers is flawed 44 I am a problem solver 45 Anthropomorphising mathematics or the universe as being female 46 Planet Contrary 47 Extreme mathematics 48 Brainstormed 49 Call for a press release 50 Tiger Woods analogy === Subject: Re: JSH: Tired of the deluded [...] I will claim on `idea man' as well. [...] What? No `crucial.' -- Michael Press === Subject: Re: JSH: Tired of the deluded posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) An excellent list - you have been paying a LOT of attention. Here are a few others 51 'Nifty result' 52 Reference to the guy in high school who did worse on the SAT 53 It's easy algebra 54 Lapdogs ... Ullrich had racist thoughts ... 55 Generals LIKE me ... 56 Your species ... 57 ... passed peer review ... 58 Extreme mathematics ... 59 A proof cannot be wrong ... proofs cannot duel 60 Perp walks ... angry mobs ... 61 Suing Harvard for its endowment ... Marcus. === Subject: Re: JSH: Tired of the deluded posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) I love the idea of this game. Sing it!!!!! === Subject: Re: JSH: Tired of the deluded I love the idea of this game. Sing it!!!!! 62. no negative sqrt(4) 63. Random numbers from k mod 3 64. Zombie from mercury in my teeth 65. HUGE 66. YEARS and YEARS 67. the Cows..... 68. The Hive 69. The Parasites 70. My research 71. easy crap that should NOT BE FREAKING NEW!!! 72. they rejected my paper === Subject: Re: JSH: Tired of the deluded posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) 73. killing the journal 74. I talked about it 75. They don't even bother 76. Prime counts 77. You people are stupid 78. I've corrected my mistakes 79. they can justify ignoring my work 80. I found a very annoying way to solve Fermat's Last Theorem 81. It's almost funny in retrospect 82. Do you really care about the truth 83. is the human search for truth a bunch of bull in the area of pure mathematics 84. I found myself musing 85. rather simple, but highly creative way 86. I may have been too smart for my own good 87. simple ideas here are apparently so brilliant that mathematicians discount them 88. a major idea which apparently eluded mathematicians 89. link between the prime number distribution and continuous functions 90. I had something that human beings had never seen before Okay - ten more for the magic 100! === Subject: Re: JSH: Tired of the deluded posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) Only 9 more if you count Yup. 92. partial difference equation 93. The reality is that I'm far ahead of mathematicians at every level 94. I'm a discoverer 95. That's only with the pure math implementation. 96. I like being the revolutionary 97. I find it odd that there has been little response to my refutations of various bogus claims made by sci.math posters 98. I have a B.Sc. in physics 99. It's impossible for something to both be the same and different. 100.I do understand it. The point is that other people do not. === Subject: Re: JSH: Tired of the deluded posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) ... and how can we forget - 101. OOPS! Marcus. === Subject: Re: JSH: Tired of the deluded posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc How indeed. Also: 102. By the RULES === Subject: Re: JSH: Tired of the deluded posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) 103. I thought some more about when that approach might work 104. You can easily find 105. The answer comes from 106. and it may work, but if it does NOT work 107. then it seems reasonable 108. and it doesn't seem likely that it can be much smaller 109. I'd think factoring should occur with a high probability 110. After years of acting as if expressions like 111. Like on planet Contrary philosophers recently decided it was immoral 112. but the people on planet Contrary have one upped you 113. we'd just resolve the freaking square root 114. Like they can prove to you 115. Recently things are restless though 116. one person has dared to stand up to the philosopher overlords 117. Get it, yet? 118. Has anything seeped through the wall of stupidity 119. may be protecting you from basic knowledge about mathematics 120. I say it's 5 or -5 and can show you (-5)(-5) = 25. 121. What say you in reply? 122. Hey, you calling me daft is like a caveman laughing at a physicist. 123. And all the definitions in the world would not save you. 124. How mathematicians screwed up is they got to sqrt(2) and stopped. 125. Dude I'm wondering how in the world people like you think you have a brain. 126. I think it's crucial to explaining why you'd play dumb on these points 127. You're too old in mind, whether you're actually too old or not. 128. You can't get started again with corrected mathematics. 129. So you're rather call me a troll than accept the truth 130. I think you'd rather die in error believing you were brilliant in your youth Can we get to 200? I almost pee'd in my pants reading the list so far! Instead of Dilbert - can we create Jamesbert? === Subject: Re: JSH: Tired of the deluded posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) Jimbert would be better (gotta be 3 letters before the -bert). === Subject: Re: JSH: Tired of the deluded posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX 5.0),gzip(gfe),gzip(gfe) Good point! So, we can have: Twelve years of delusion and narcisscism and still counting: Jimbert the series! === Subject: Re: JSH: Tired of the deluded posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) A lot of the Dilbert books can be Jimbert titles with just slight alteration: Thriving On Vague Object Rings Random Acts of Mathematics It's Not Funny If I Have To Explain It* Always Postpone Demonstrating Examples To Timewasting Mathematicians Bring Me The Head Of David Ullrich I'm Not Anti-Mathematics, I'm An Idiot Don't Step In The Galois Theory When Usenet Goes Bad It's Obvious You Won't Survive On Your B.Sc. Alone When Did Ignorance Become A Point Of View?* What Do You Call A Sociopath On Usenet: A Discoverer *- no alteration required === Subject: Re: JSH: Tired of the deluded posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) 91. You're lying. 92. Nonpolynomial factorization ... 93. Games theory ... 94. The constant term is constant. 95. a Magidin mathematician ... 96. a BF Circle ... 97. Barry Mazur ... Andrew Granville ... Odlyzko ... 98. Ralph McKenzie 99. The discoverer eventually wins. 100. ... my partial differential ... Bingo! Marcus. === Subject: Re: JSH: Tired of the deluded posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 4334.34; Windows NT 5.1; SV1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) spider-dtc-tc08.proxy.aol.com[CDBC7048] (Prism/1.2.1), HTTP/1.1 cache-dtc-ad05.proxy.aol.com[CDBC74C7] (Traffic-Server/6.1.5 [uScM]) Shouldn't yup be on the list? Maybe a 5x5 grid isn't big enough. Or maybe play Blackout, where the object is to mark EVERY square. === Subject: Re: JSH: Tired of the deluded posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) You delusional narcissistic moron - your method doesn't even work! === Subject: boook posting-account=pgGZQwoAAAByAwM34BOS4uYIOEnRoHog Basic Principles and Calculations in Chemical Engineering 7th edition I am wondering if i can get a copy of it! can u email me at paulhuynh2o(at)Gmail.com thank u === Subject: Re: boook Yes, of course you can. It is in print (ISBN-10: 0131233688, ISBN-13: 978-0131233683) and will therefore be available from book shops either directly or to order. Why are you asking about a chemical engineering book in a mathematics newsgroup? Why should I? -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. === Subject: Solution Req. posting-account=0Bc2OwoAAABK_rp6RBnkr560GLyKsAIz 1.1.4322; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30),gzip(gfe),gzip(gfe) i need solution of this problem: A market stall holder buys baskets of vegetables at Rs. 2 per Kg. he sells each Kg for Rs. 5, but there is a probability of 1/18 that a basket is unfit for sale and he will lose Rs.2 per Kg.find the average profit. Umar === Subject: Re: Solution Req. Average profit = (probability that basket is fit) * (profit when basket is fit) + (probability that basket is unfit) * (profit when basket is unfit). You fill in the details. (Remember that a loss is a negative profit.) HTH -- --------------------------- | BBB b Barbara at LivingHistory stop co stop uk | B B aa rrr b | | BBB a a r bbb | Quidquid latine dictum sit, | B B a a r b b | altum viditur. | BBB aa a r bbb | ----------------------------- === Subject: Re: Solution Manual Required How can I do that? -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. === Subject: Re: Solution manual for advanced engineering mathematics Wylie You'll get more help from the publisher. -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. === Subject: Re: Solutions If it exists, yes, otherwise no. -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. === Subject: Re: Internet Calculations - calculate.richiebrownlee.com http://en.wikipedia.org/wiki/Phasor_%28sine_waves%29, maybe. -- Which of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. === Subject: Solution to Elements of Electromagnetics Mathew O Sadiku 3rd edition posting-account=I8pLeQoAAACEQHjwkZDGQuLi6GDhaS5D 5.1),gzip(gfe),gzip(gfe) I have solutions of few chapters (from CH3 to CH11)Elements of Electromagnetics Mathew O Sadiku 3rd edition its worth it..if anyone needs it contact me ! its for $15..! === Subject: Re: Solution to Elements of Electromagnetics Mathew O Sadiku 3rd edition posting-account=SJsYBwoAAACWOFqNtTFT_QkIWPHz2DA6 wats ur email id..? i cant see it.. n' im interested in the solutions i found a copy for free online.. is tat the same one u have.. or do u have a instructors manual? === Subject: Re: instructor's solution manual for Elements of Electromagnetics 4th ed posting-account=I8pLeQoAAACEQHjwkZDGQuLi6GDhaS5D 5.1),gzip(gfe),gzip(gfe) i have it ffrom CH3 to CH11 (odd+even) ..if you want it snd $15 to eerye887@gmail.com === Subject: Solution of Introduction to Algorithms second editionCormen ,Leiserson posting-account=I8pLeQoAAACEQHjwkZDGQuLi6GDhaS5D 5.1),gzip(gfe),gzip(gfe) i have solution manual of Solution of Introduction to Algorithms second editionCormen ,Leiserson snd me $15 through paypal to eerye887@gmail.com .and il snd u the solution right away. === Subject: Solution manual of Advanced Engineering Mathematics Dennis G.Zill Michael R.Cullen 3rd edition posting-account=I8pLeQoAAACEQHjwkZDGQuLi6GDhaS5D 5.1),gzip(gfe),gzip(gfe) i have complete solution manual Solution manual of Advanced Engineering Mathematics Dennis G.Zill Michael R.Cullen 3rd edition snd me $15 through paypal eerye887@gmail.com and il snd the link === Subject: Re: Why is i squared always -1? EXPLAIN PLEASE SHH! We don't want the public to know that! === Subject: Re: Why is i squared always -1? EXPLAIN PLEASE Yes, you are right. That is, however, what I meant with my last line. The whole point of defining the set of complex numbers to be the collection of pairs of real numbers, and (a, b)(c, d)= (ac- bd, ad+ bc) is to define something whose square is -1. === Subject: Re: Solutions manual for Machine Design: An Integrated Approach, 3rd Ed., by Robert L. Norton Hey, I am taking a machine design class with the same textbook you are using. Can you please send me the solutions manual for the book: Machine Design: An Integrated Approach, 3rd Edition. bwk5011@psu.edu === Subject: Re: Solutions manual for Machine Design: An Integrated Approach, 3rd Ed., fishimishi@gmail.com === Subject: Question on set membership Let us consider a family of semi-open intervals A = { [(1/2^(n+1)) , (1/2^n) ) : n = 0,1,2,3 .... } just to make things clear I will enumerate few of these intervals A = { [0.5, 1), [0.25, 0.5) , [0.125, 0.25) ...... } Can someone please explain why {0} is not a member of A ? === Subject: Re: Question on set membership doesn't make much sense - {0} is an element of A or it is not, period. Now, you've given an enumeration of the elements of A: the elements are [1/2, 1/4), [1/4, 1/8), [1/8. 1/16), etc. Any set on that list is an element of A, and every element of A is on that list. So. Which set on that list is equal to {0}? (Is {0} = [1/2. 1/4)? No. Is {0} = [1/4, 1/8)? No. . . .) David C. Ullrich === Subject: Re: Question on set membership On Sat, 02 Feb 2008 13:55:15 EST, convoluted in alt.math.undergrad: The elements of A are precisely the sets of the form [1/2^(n+1), 1/2^n); the set {0} is not of that form, so it's not an element of A. Or were you actually wondering why 0 is not a member of the *union* of A? That's true because 0 is not an element of any member of A: no matter what n in N you choose, 0 is not an element of [1/2^(n+1), 1/2^n). Brian === Subject: Re: 2000 Solutions manual posting-account=dvG8ZwkAAABquTxNLH5eYgLcRJ04av3D I would need: A brief introduction to fluid mechanics: student solutions manual. (the 3rd and the 4th if possible) it would be awesome. === Subject: Re: Complete Solutions Manuals For College and Graduate Textbooks posting-account=vYfegAoAAACyHTb9CEn_lMUMQdyLpp5z Hii can i get solution manual of SIGNALS SYSTEMS AND TRANSFORMS 3rd edition by Charles Prentice I will appreciate ur help Raghav === Subject: Re: 2500 Availables solutions manuals posting-account=h7dUlgoAAACqhfPqhmy6NNx8o2svcdeS I just sent you an email regarding the Sol. Manuals. Do you have the manual for Fundamentals of Structural Dynamics By Craig? Do I get any discounts if I purchase multiple items? === Subject: Re: 2500 Availables solutions manuals posting-account=Iy9uCgoAAAA2PU5tzAhkCUIzJxrSNXRJ 1.1.4322),gzip(gfe),gzip(gfe) I need the solution manual of the book Mechanical Engineering Design (6th Ed., Shigley). Please, give all the information, as price and the way of payment. I want to buy it. Do you have the solution manual of the book heat transfer any edition, Hollman? === Subject: Re: perfect square posting-account=q2ruDwoAAACW4OvF4soI61szUIc6xJHw Problem has been resolved by recent posting to sci.math, with subject = 2^2 + 2^11 + 2^n === Subject: Solution Manual of Engineering Mechanics - Statics And Dynamics 11ed by Hibbeler in pdf form posting-account=maF0DwoAAABHu3c_lYtWkivYbsdUjqO2 I have the solution manual in pdf form of the book:Engineering Mechanics - Statics And Dynamics 11ed by Hibbeler. It have all the solution to each and every question in the book worked out in very easy language. Here are some details of the solution manual Book: Engineering Mechanics - Statics And Dynamics Edition: 11th ( it is the latest edition) Author:Hibbeler Language: Simple English This solution is in pdf form and will be sent through email to your paypal address or the address specified by you. The language is simple with each and every question worked out in very very simple way which is very easy to understand and score high grades. if you have any question feel free to ask me. can also contact me at nikhilpali_88@yahoo.com === Subject: solution manuals may i have all solution manuals that have ever been produced even if they've got nothing to do with undergraduate mathematics email them to me now hey yeah and plz -- Going forward at this moment in time a raft of measures have been put in place on the ground to target and claw back the growth of cliche usage 24/7. === Subject: quick question Hello All, Can anyone advise (provide a solid yet simple argument) as to why the numbers 81^n and 16^n always end in 1 and 6 respectively for any n element of the natural numbers. Best, George === Subject: Re: quick question 81^n = (80+1)^n = 80^n + ... + n*80*1^(n-1) + 1^n Since every term of that expansion, except the last, contains 80 as a factor, every term except the last contains 10 as a factor. That means that the last digit (units digit) of 81^n is the same as the last digit (only digit) of the 1^n term, which is 1. 16^n = (10+6)^n can be analyzed in a similar way. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ If there's one thing I know, it's men. I ought to: it's been my life work. -- Marie Dressler, in /Dinner at Eight/ === Subject: Re: quick question alt.math.undergrad: First realize that a positive integer ends in 1 (when written in the usual decimal notation) precisely when it is one more than a multiple of ten, i.e., of the form 10k + 1 for some (non-negative) integer k. Similarly, it ends in 6 precisely when it's of the form 10k + 6. Suppose that for some particular value of n, 81^n just happens to end in 1. This means that 81^n is one more than a multiple of 10: 81^n = 10k + 1 for some integer k. But then 81^(n+1) = 81 * 81^n = 81 * (10k + 1) = 810k + 81 = 810k + 80 + 1 = 10 * (81k + 8) + 1; since this is one more than a multiple of 10, it must end in 1 when written in the usual decimal notation. argument just given, 81^(1+1), i.e., 81^2, also ends in 1. But then 81^3 = 81^(2+1) ends in 1, and hence 81^4 = 81^(3+1) ends in 1, and so on. A very similar argument works with 16^n; the key calculation is 16^(n+1) = 16 * 16^n = 16 * (10k + 6) = 160k + 96 = 160k + 90 + 6 = 10 * (16k + 9) + 6. The technical term for such an argument is 'proof by (mathematical) induction. Brian === Subject: Re: quick question binomial expansion on (80+1)^n and (10+6)^n and hoping something would click. I never considered induction. === Subject: JSH: Little congruence result posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif With non-zero coprime integers n_1 and n_2, if f_1 = r_1 mod n_1 and f_1 = r_2 mod n_2, you can find f_1 mod p_1*p_2 with f_1 = r_1 + j*n_1 mod n_1*n_2 where j = (r_2 - r_1)*n_1^{-1} mod n_2. I like this result better than the Chinese Remainder theorem, as you can just go by two's--and it's my discovery. I like primarily using my own math. James Harris === Subject: Re: JSH: Little congruence result posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Awesome little result where I find it amazing that no one has found it before as to me it's so much simpler to understand and use than the Chinese Remainder theorem, like notice, it's quite obvious and direct why n_1 MUST be coprime to n_2, as you need its modular inverse modulo n_2. And you can use my little congruence result with as many composites as you want and figure out immediately how many iterations it will take, as if n is the number of composites then ln n/ln 2 is the approximate number. So to do 128 composites would require 7 iterations, taking your results by two's. That little tidbit is just an extra though for this latest research path. I like those little extras. Last little extra for me was my prime counting function coming in after I discovered my proof of Fermat's Last Theorem. These tidbits are interesting in that they cover areas where mathematicians already have something, where I just find something better. James Harris === Subject: Re: Little congruence result So you are saying If f_1 = r_1 mod n_1 = r_2 mod n_2, then f_1 = r_1 + j*n_1 mod n_1*n_2 where j = (r_2 - r_1)*n_1^{-1} mod n_2. Or f_1 = r_1 mod n_1 = r_2 mod n_2 = r_1 + ((r_2 - r_1)*n_1^{-1} mod n_2) * (n_1 mod n_1*n_2) How can that be ? === Subject: Vector Mechanics for Engineers, Dynamics posting-account=NIP6TQoAAAABzEn6zNfq-kXNT7SBNjNB Gecko/20061201 Firefox/2.0.0.11 (Ubuntu-feisty),gzip(gfe),gzip(gfe) I have the solution manual (pdf) Vector Mechanics for Engineers, Dynamics 7th edition. Im selling it for $18. The only form of payment ill accept is paypal. email me if interested. at getreadytosucceed@gmail.com === Subject: Mechanics of Materials (6th Edition) by Russell C. Hibbeler posting-account=NIP6TQoAAAABzEn6zNfq-kXNT7SBNjNB Gecko/20061201 Firefox/2.0.0.11 (Ubuntu-feisty),gzip(gfe),gzip(gfe) Im selling the solutions manual(pdf) for Mechanics of Materials (6th Edition) by Russell C. Hibbeler for $15. the only form of payment is $15. email me if interested getreadytosucceed@gmail.com === Subject: Re: Vector Mechanics for Engineers: Statics (8th Ed., Ferdinand P. Beer) so posting-account=BoX_1AoAAAClnbnLjCtTfuXZBDivZvL7 hey guys it would be of great help if i could get the solutions manual. please? email me at zbarhoumi@gmail.com === Subject: =?windows-1256?B?ys7h1e0g4+QgytTe3iDD3s/H498gyMfh1NHNIA==?= =?windows-1256?B?x+Hmx93tIMjH4dXm0Q==?= posting-account=OEMLPAoAAAD96Le1WgD9Upxh2I266gSV SIMBAR={94AA8A13-642B-404A-B4A4-C4EFD1151A5B}; Alexa Toolbar),gzip(gfe),gzip(gfe) === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif LOTS!!! Wrong idea. Don't bother considering it. Just one of those things. Another mistake of mine. Nothing of consequence. Please move along...nothing to see here. James Harris === Subject: Re: Solution to the factoring problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) But it's easy algebra, right? Marcus. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif Yup! And I screwed it up. But I got my little congruence result out of it. Mistakes are part of mathematical discovery. Every major result I have has been preceded by lots of stupid mistakes. And I mean REALLY stupid mistakes. So making stupid mistakes tells me that I'm in pursuit of something. When I just get everything right all along then it's a nothing result. This one is HUGE. With the little congruence result, you can do so many things... And who else has replaced the Chinese Remainder theorem? James Harris === Subject: Re: Solution to the factoring problem posting-account=sFP0HgkAAADJMwhdrXAaC5VX7Tc3BtzY 1.0.3705; .NET CLR 1.1.4322; .NET CLR 2.0.50727; Media Center PC 4.0),gzip(gfe),gzip(gfe) Great, you got a trivial one-liner out of days of ranting and making a clown out of yourself again. You think Gauss would agree with you on that too? Dead-end ideas are part of mathematical discovery, not blunders of the JSH format. Every mathematical statement you have uttered has been preceded (and followed) by lots of stupid mistakes. Yeah, that's what it tells you. It would tell most people they need to study it better before next time they announce to the world they are the greatest genius in that field ever. Also then. It's trivial. --- J K Haugland http://home.no.net/zamunda === Subject: Re: Solution to the factoring problem posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc But you're still a genius, right? And the people who told you you screwed it up are still parasites, is that it? Replaced it with a trivial special case? I have, for one, within about a minute of reading the statement of the theorem on Wikipedia, since proving the case with two residues is a piece of piss. Most people don't feel the need to announce it to the world when they rederive special cases of results that were known 17 centuries ago. === Subject: Re: Solution to the factoring problem posting-account=Ae7cPwoAAAA1p9Bl1szxMYHINqjRfucA CLR 2.0.50727),gzip(gfe),gzip(gfe) Those implementing should follow the protocol I mentioned earlier where groups post answers in concert so as to prevent attempts at suppression of the actual factors of T mod p_1 such that T/m! < 1. Then you do the same for p_2, and should have the order as well. And you just do that for a target T is take two primes p_1 and p_2, and should have the same with p_1*p_2, finding g_1 and g_2 such that f_1*f_2 = T mod p_1 and f_2 mod 11 = 6. Those are just numbers. Just answers and then loop through those results go modulo your first prime and then store all values again for factors of T. And you do the same modulo p_1*p_2 you get answers, which are, notice 7 mod 11 and f_2 mod p_1, and you have enough answers and then match back to your first prime and compare back with your first list, and you have the correct answer for your factors but which one is correct? To find it you loop through those results go modulo your first list, and you have 7 mod 13 = 4. Can you see the difference? That is why when you move modulo p_1*p_2 and the right answer is in there somewhere. Do the same set, so now you have the same for p_2, and for f_2 mod p and f_2 mod p and f_2 = T mod p_1, and those are the residues modulo p_1 of the information by unethical governments. I've read through your reply. And I told you I checked it. It can be generalized, but the target composite to be factored that is the correct one must be in the group. Then you loop through all possible factors f_1 mod p_1*p_2 and then use the Chinese Remainder theorem to get f_1 mod p_key, and then you should be able to factor an RSA public key. The factoring problem is solved and simply solved where for some reason mathematicians just missed it. But it does solve the factoring problem. A large overestimate for the number of primes needed is m such that f_1*f_2 = T mod p_next, which is true, but it does solve the factoring problem. A large overestimate for the number of primes needed is m such that f_1*f_2 = 119 mod 11. And that is 9 mod 11, so you'd have f_1 mod p and then maybe you'll be able to show it with the mathematics versus insulting me or just telling people I'm wrong. I wish to remind that I'm dealing with parasitic beings who are now caught in a bind with a solution previously given for p_1, and find f_1 mod p_2 and f_2 mod p and f_2 mod p and f_2 mod p_1, as you can use it all the values for p_key as you loop back through that list modulo each prime in turn, and find f_1 mod p_1 such that T/m! < 1. So even the largest RSA public key currently in use or practically useable. Think that it is true that if f_1 = r mod p_1. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif It can be generalized, but the target now is RSA encryption. They can be any primes, but it seems useless to use small primes. I'd think that starting with 101 and going up would be sufficient. Positive residues, so of course they are integers. Consider my favorite example T = 7(17)=119, with p=11. For this technique assuming you don't know the factors you'd get all possible factors f_1 mod 11 and f_2 mod 11 such that f_1*f_2 = 119 mod 11. And that is 9 mod 11, so you'd have: 1, 9 2, 10 3, 3 and so forth Like now you could have p_2 = 13 to go along with p_1 = 11, so you'd look for 119 mod 143, which gives you a list of 142 numbers, again starting with 1: 1, 119 and so forth Now then in each of those lists you must have the correct answer for your factors but which one is correct? To find it you loop back through the second list, taken mod 143 and take the solution modulo 11, and then match back to your first list. You can go prime by prime finding f_1 mod p, so you'd have f_1 mod p_1 f_1 mod p_2 f_1 mod p_2 and so forth, so you can use the Chinese Remainder theorem to find f_1 mod p_1*p_2*p_2*...*p_j where you have enough primes that f_1 equals your factor. j is less than m such that T/m! < 1. So you'd never go over 200 primes even to factor the largest RSA public key currently in use or practically useable. Think that it is easy and then maybe you'll be able to understand it as it IS easy. Oh, it's easy. With a target T, you can find approximately how many primes you will need using m such that T/m! < 1. Then you start with what I call the key prime, which I think could just be 101. You take T mod p_key, and store all the values for f_1 mod p_key and f_2 mod p_key such that f_1*f_2 = T mod p_key. Then you get your next prime p_next and multiply times p_key, and then store all values again for factors of T mod p_key*p_next. Then you loop through those solutions taking each in turn modulo p_key, and comparing with your first set of stored values, pulling all matches. One of those is the correct value for f_1 mod p_key, and for f_2 mod p_key. You may get more than one value depending on how many prime factors T has, and whether or not you get negative solutions. You now also get f_1 mod p_next and f_2 mod p_next by checking your possibles by taking the original results--not the mod p_key ones-- modulo p_next, and multiplying them together and seeing if you get T mod p_next, which is a way to resolve multiple solutions. And now you have f_1 mod p_key, and f_1 mod p_next. So you go to the next prime and repeat, maintaining the stored values for p_key as you can use it all the way down. You do this until you have enough answers and then use the Chinese Remainder theorem to get f_1. Properly implemented it should be faster than an RSA key can be generated, no matter what the size of the key. So you'll be able to factor an RSA public key faster than a computer can even generate the thing, no matter what its size. James Harris === Subject: Re: Solution to the factoring problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) say, I loop back through the second list, taken mod p_1*p_2, I am going through a truly enormous list. This *may* give me the right answer, but it certainly does not appear to be efficient. Why would you necessarily arrive at the correct factor this way? For example, with T = 1189, if I choose p_1 = 37, then since T mod 37 = 5, I get f_1 = 2 and f_2 = 21. I then choose p_2 = 43 and I find that 26 * 11 mod p_2 = 28 mod p_2 = T mod p_2. So I let f_2 = 26. I have two congruences, f_1 mod p_1 = 2 mod 37 and f_1 mod p_2 = 26 mod 43, and I use the CRT to find the smallest postive value of f_1 which satisfies both of these congruences, namely, f_1 = 1445. But this is not a factor of T = 1189. And if I add a third prime p_3, the value I will find for f_1 is going to be even bigger. That is, this method will not lead to a factor of T. Marcus. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif That's not my advice. In the past with the factoring congruences I p will give the smallest prime factor if they are non-trivial and T has only two prime factors. With the solution to the factoring problem, however, an overestimate of the number of primes needed is given by m, where T/m! < 1. Between 100 and 1000 there are 143 primes, so this approach can handle composites as large as 143!. The approach is amazingly simple as you just take two primes at a time, and loop through finding f_1 mod p_1 and f_2 mod p_1, such that f_1*f_2 = T mod p_1 and then do the same modulo p_1*p_2, so that you have g_1*g_2 = T mod p_1*p_2 and then loop through those results modulo p_1, and note cases where you have an intersection. Those are the cases where you have factors of T. It's a trivially simple idea which relies on the fact that f_1 mod p_1*p_2 = f_1 mod p_1 so there's no way to debate it. The fact that it wasn't known before may be partly about how modern mathematicians do their research as they like to build on past research, learning how previous mathematicians approached a problem looking to build versus innovate. But back in the times of say Gauss or Euler, this method would have been tedious and undesirable because of the mindlessness and simplicity of it. I'm sure Gauss would have disdained it for being too...easy and tedious. But for modern computers it's perfect. Mathematicians failed because they forgot that creativity and mathematical discovery is not about technique but about magic. That's why they call it genius. There is a genie in the equations. And the genie can never be completely contained by human minds, wants or wishes. James Harris === Subject: Re: Solution to the factoring problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) Nothing like an example, eh? Let T = 1189, and p_1 = 7, p_2 = 11, p_3 = 13, p_4 = 19. Then you get the following little table (where in each case, f_1 * f_2 = T mod p_i): T p_i T modp_i f_1 modp_i f_2 modp_i ---- ---- --------- ---------- ---------- 1189 7 6 4 5 1189 11 1 6 2 1189 13 6 6 1 1189 19 11 10 3 So then you use the Chinese Remainder Theorem to find the smallest positive f so that f mod p_1 = f_1 mod p_1 f mod p_2 = f_1 mod p_2 f mod p_3 = f_1 mod p_3 f mod p_4 = f_1 mod p_4 That f turns out to be f = 6584, which is clearly not a divisor of T = 1189. In fact it is coprime to 1189. If you do the same thing for just the first three of these primes (7, 11, 13), you get f = 578 = 2*17^2, which is also coprime to T. Conclusion: Your method does not produce a factorization of T. Marcus. === Subject: Re: Solution to the factoring problem posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) [.snip] Delusional narcissist - you have nothing! === Subject: Re: JSH: Solution to the factoring problem [added JSH: to subject] James, if g_1*g_2 = T mod p_1*p_2 [1] then these are also true: g_1*g_2 = T mod p_1 [2] and g_1*g_2 = T mod p_2 [3] This is an elementary property of congruences -- it doesn't even matter whether p_1 and p_2 are prime or composite. As a result, /every/ pair you find satisfying [1] passes the [2] and [3] tests. Your intersection is the entire set of solutions to [1]. I know you claimed to have worked out an example before, but you didn't show your work. If you were telling the truth about working out an example, then you made arithmetic errors. Here's a complete example without arithmetic errors: T = 11*13, p_1=3, p_2 =5. These are all the solutions to i*j = 11*13 (modulo 3*5) [4] with 0 <= i <= j < 3*5: i j - - 1 8 2 4 7 14 11 13 For example, 7*14 = 98 = 8 = 11*13 (modulo 15), and it's also the case that 7*14 = 11*13 (modulo 3) and that 7*14 = 11*13 (modulo 5) them modulo 3 or modulo 5. This is always the case, regardless of the values of T, p_1, p_2, g_1 and g_2. === Subject: Re: JSH: Solution to the factoring problem posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Actually I think that James means for his method to work the other way round, that is to say you use solutions to [1] to eliminate solutions to [2] and [3]. But this doesn't work either, since it is easily seen that the projections that map solutions to [1] to solutions to [2] and [3] are surjective. In fact, the CRT shows that checking the intersection of solutions to [1] with those to [2] and [3] doesn't even eliminate any /quadruples/ (g1,g2,h1,h2) with g1,g2 satisfying [2] and h1,h2 satisfying [3]. So in the end this method will mean trying all possible sequences of residues fi mod pi, with pi a product behaviour as trial division even if you assume that checking each sequence takes O(1) steps. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif That's wrong. I've checked it already with simple examples and it does give the factors modulo the particular primes and it's such a brilliantly simple idea I've been puzzling why no one figured it out before now. What you do for a target T is take two primes p_1 and p_2, where first you take T mod p_1, and find f_1 mod p_1 and f_2 mod p_1 such that f_1*f_2 = T mod p_1 So for instance the start would be f_1 = 1 mod p_1, and f_2 = T mod p_1, and the second would be f_1 = 2 mod p_2 and f_2 = 2^{-1}T mod p_1, and you go down the line finding every possible residue modulo p_1 for your factors. So the correct one must be in that group. Then you do the same with p_1*p_2, finding g_1 and g_2 such that g_1*g_2 = T mod p_1*p_2 and go down the line again, getting EVERY possible, so the correct answer MUST be in the group. Then you go back through that list modulo p_1 and match back to your first list, and you have the correct f_1 mod p_1 and f_2 mod p_1, and then you also get f_1 mod p_2 and f_2 mod p_2 by taking the previous modulo p_2. One thing that can happen is that you get two possibles because negative residues can work, but that's easily handled. Why does this work? Because for integer f_1 and f_2 such that f_1*f_2 = T, each has some residue modulo p_1, as well as a residue modulo p_1*p_2, BUT it is true that if f_1 = r mod p_1*p_2, then f_1 = r mod p_1. Puzzling over how this easy technique is new my guess is that neither Gauss nor Euler would like it as it is very tedious to do by hand, while it's perfect for modern computers. Modern mathematicians tend to start first by considering what was done before and try to add on, not innovate or find something totally new from scratch, so somehow they just missed it. But it does solve the factoring problem. A large overestimate for the number of primes needed is m such that T/ m! < 1. So even the largest RSA public key would be handled by I'm sure less than 200 primes. If you consider primes near 1000 where I remind there are 168 up to 1000, then the largest search space if p_1 = 101, and you used a prime very close to 1000 would be about 101000. So the time would be roughly 168*1000 iterations to handle even the largest RSA public key. The factoring problem is solved and simply solved where for some reason mathematicians just missed a fairly trivial solution. Those implementing should follow the protocol I mentioned earlier where groups post answers in concert so as to prevent attempts at suppression of the information by unethical governments. James Harris === Subject: Re: Solution to the factoring problem posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc It doesn't work, as explained in my second reply. I will do so again below, in more detail. I assume you are familiar with the fact that any integer f has an inverse mod N iff f and N are coprime. Yes, but that group contains *every* non-zero f1 mod p1. This is because any f1 chosen from {1,2,...p1 - 1} is coprime to p1, and therefore has an inverse mod p1. So we may take f2 = T*f1^(-1) mod p1, and this satisfies f1*f2 = T mod p1. Any g1 in {1,2,...p1*p2 -1} which is not divisible by p1 or p2 is coprime to p1*p2 and therefore has an inverse mod p1*p2, so any such g1 lies in this group. No, you have any f1, not just the correct ones. For consider any f1 in {1,2,...p1 - 1}: if f1 is not divisible by p2, let g1 = f1. Clearly g1 is not divisible by p1, so from the above g1 has an inverse mod p1*p2, so there exists g2 such that g1*g2 = T mod p1*p2, and this g1 is equal mod p1 to f1. If on the other hand f1 is divisible by p2, then let g1 = f1 + p1. This is not divisible by p2 since p2 divides f1 but not p1, nor is it divisible by p1 since p1 divides p1 but not f1. So from the above g1 has a modular inverse mod p1*p2, and so there exists g2 such that g1*g2 = T mod p1*p2, and this g1 is equal mod p1 to f1. So by comparing the lists of possible values for f1 mod p1 and g1 mod p1*p2 narrows the possible values for f1 mod p1 down to... all of them. Note that comparing values of f2 mod p1 and g2 mod p1*p2 won't help you here either; since the integers mod p1 are a field, to any f1 != 0 mod p there is a *unique* f2 mod p1 such that f1*f2 = T mod p1, therefore the values of g2 found above will necessarily satisfy g2 = f2 mod p1, and comparing them tells you nothing. Right. But this doesn't help, since it's not true that f1*f2 = T mod p1, g1*g2 = T mod p1*p2 and g1 = f1 mod p1 together imply that f1 is equal mod p1 to an actual factor of T. === Subject: Re: Solution to the factoring problem It is always risky to assume that JSH is aware of any mathematical facts, even when he appears to be relying on them. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif I challenge any posters replying here to just try an example. If you're right then you should be able to show it with the mathematics versus insulting me or just telling people I'm wrong. I wish to remind that I'm dealing with parasitic beings who are now caught in a bind with a simple mathematical result. They can't answer with an example and will not give up control easily. Like any parasites, they will probably require forcible removal. James Harris === Subject: Re: Solution to the factoring problem posting-account=BVr-MgkAAABE4LRE1rHDnN9heo0IZZTk .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) spider-ntc-ta11.proxy.aol.com[CFC8700B] (Prism/1.2.1), HTTP/1.1 cache-ntc-ab05.proxy.aol.com[CFC87445] (Traffic-Server/6.1.5 [uScM]) http://forums.worldofwarcraft.com/thread.html;jsessionid=3D3935719DB6BF60119 08E270931C40A?topicId=3773504246&sid=1 Enrico === Subject: Re: Solution to the factoring problem posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) Where is your worked example - delusional narcissist? === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif I've read through your reply. And I told you I checked it. What you need to do is just try it. It seems you're lost on the result you mention above which is true, but it does not change the reality that given integers f_1 and f_2 f_1 mod p is one value, as is f_2 mod p if you take the positive residues. For instance, with 7(17) = 119, 7 mod 11 = 7, and 17 mod 11 = 6. Those are just numbers. Just answers and they don't shift. So now if you take your second prime as 13, notice that again you have 7 mod 13 = 7, but now you have 17 mod 13 = 4. Can you see the difference? That is why when you move modulo p_1*p_2 you get answers, which are, notice 7 mod 143 = 7, and 17 mod 143 = 17. So again there is a shift. Try the technique with p_1 = 5 and p_2 = 11, and T=119. That's easy so you have no excuse not to see what happens. I already know as I've done it. It works. So your thinking here is totally wrong. I don't want you clogging up my threads with deluded crap though so try the example before you start replying all over the place with some kind of religious zealotry that people like you display. The time and place for that b.s. has gone. James Harris === Subject: Re: Solution to the factoring problem posting-account=wVv_VwoAAAAVTfUuyxLzug5SzYWCgHj1 Gecko/20061025 Firefox/1.5.0.8,gzip(gfe),gzip(gfe) The example you gave above is not useful, because you started with f_1 and f_2 the factors of T. You need to start with candidate factors mod p that are not necessarily the true factors, then show how you are led to to the true factors. Here is an example. Let T = 1189. Let p = 37. Then T mod p = 1189 mod 37 = 5. I choose f_1 = 2 and f_2 = 21, and it is true that f_1 * f_2 mod p = 2 * 21 mod 37 = 5. So now I choose a second prime, say, p = 43. I note that f_1 mod 43 = 2 and f_2 mod 43 = 21, just as before. And of course, f_1 mod 37*43 = 2 and f_2 mod 37*43 = 21. If I keep going in this way, with larger values for the primes, I will never get any answers other than f_1 = 2 and f_2 = 21. This does not lead me to the correct factorization. You need to show how your method works with, for example, T = 1189. I must note here also that your method, which you said was proven to work by 'simple algebra', keeps shifting and changing. If you have to consider all possible factorizations mod p, where p is a large prime, you are talking about a lot of calculation, which gets back to the fact that you don't really have a proof that your method (whatever your current method is) is efficient. Marcus. === Subject: Re: Solution to the factoring problem posting-account=aLpfCwoAAACh4BOs3HOlQBCoxUpEgyxc Good advice. Yes, but the problem is that *we* *don't* *know* *what* *the* *values* *are* until we know the factorisation of T. So what? You got those numbers by simply calculating the residues mod p of factors of T. What do you do if you don't know the factors of T? You have the problem completely backwards. Try what technique? What am I supposed to do if I don't know that 119 factors as 7*17? How about this example, again with p1 = 11 and p2 = 13: try f1 = 2, f2 = 131. Then f1 mod 11 = 2, f2 mod 11 = 10, which multiply together to equal 119 mod 11. Also f1 mod 13 = 2, f2 mod 13 = 1, which multiply together to equal 119 mod 13. Also f1*f2 = 119 mod 143. But this doesn't help me find a factor of 119, since f1 is not equal to a factor of 119 mod 11, mod 13 or mod 143. So how is your method supposed to work if I don't know the factorisation of T in advance? I've a better idea. You're the one who claims that this will be able to factor RSA-sized composites. YOU try it. You don't have to post a factorisation or even announce whether you succeeded if you're worried for your life. Just try it. Practice what you preach. === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif But you know they exist. So if you loop through ALL possible f_1 and f_2 modulo p_1, then you have the right answer in there somewhere. Do the same modulo p_1*p_2 and loop through ALL possible f_1 and f_2 modulo p_1*p_2 and the right answer is in there somewhere. And now use the result that f_1 mod p_1*p_2 = f_1 mod p_1, as you loop through those results and take them modulo p_1, to compare back with your original list modulo p_1, and see where you have an intersection. Easy. Assume you don't know the answers and then loop through all possible solutions modulo your first prime and then do so again modulo the product of your primes and then with those results go modulo your first prime and compare back to the first list. Huh? Weird. That's just a weird mish-mash which doesn't relate at all to the method I've been describing repeatedly throughout this thread. It's simple: with your first prime you find ALL solutions for T modulo that prime and then you multiply times your second prime and find ALL solutions modulo that product, and loop through those modulo your first prime and compare back to your first list. Where you have matches you have your answer. James Harris === Subject: Re: Solution to the factoring problem Correct, but how large is the set of ALL possible f's? If you cannot come up with a *small* set then your method will be slow. How large will the set be in the case of an RSA number? Try working out how big - it might be instructive. rossum === Subject: Re: Solution to the factoring problem posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif (p_1 - 1)/2 possibles for f_1 and f_2 such that f_1*f_2 = T mod p_1 If you use 101 as a minimum prime and just count the number of primes from 100 to 1000 you have 143 primes, which means you can handle a number at least as big as 143! within that range. If you go with the approximate size of the largest prime up to 1000, as an overestimate you have at most 143(500) = 71500. And that's for a composite at 143!. James Harris === Subject: Re: Solution to the factoring problem It seems to me the number of combinations that need to be checked is at least (p_1*p_2*...). Thus, to factor T, you would need on the order of T steps. If so, forget about trying to factor anything big. quasi === Subject: =?windows-1256?B?ys7h1e0g4+QgytTe3iDD3s/H498gyMfh1NHNIA==?= =?windows-1256?B?x+Hmx93tIMjH4dXm0Q==?= posting-account=OEMLPAoAAAD96Le1WgD9Upxh2I266gSV SIMBAR={94AA8A13-642B-404A-B4A4-C4EFD1151A5B}; Alexa Toolbar),gzip(gfe),gzip(gfe) === Subject: binomial question Why is that then when n=2k (all k in Natural Numbers) all binomial coefficients (n choose k) for k=1,2,...,n-1 [we ignore 0 and n because they === Subject: Re: binomial question ***CORRECTION: Why is that then when n=2^k (all k in Natural Numbers) all binomial coefficients (n choose k) for k=1,2,...,n-1 [we ignore 0 and n because they === Subject: Re: binomial question 6 choose 2 = 15 -- Paul Sperry Columbia, SC (USA) === Subject: Re: binomial question Ouch! I meant n=2^k. Sorry about that. === Subject: Re: binomial question alt.math.undergrad: Doesn't work if k = 1: C(2^1, 2) = 1. 2^k * (2^k - 1)/2 = 2^(k-1) * (2^k - 1), and 2^(k-1) is even Brian === Subject: Re: binomial question Let me try and state the conjecture more precisely. Apologies, I've been very sluggish in clarity. *Let X denote the set of n for which all binomial coefficients C(n,k), except the first and last, are even numbers. Then X={2^k, k=1,2,...}* So excluding the first and last coefficients tells us we are only considering coefficients which statisfy 0