61. CTK Exchange chinese remainder theorem. Yours is an example of problems solved in generalcase by what s known as the chinese remainder theorem. http://www.cut-the-knot.com/exchange/chinese2.shtml

Username: Password: Sites for teachers Sites for parents Awards Interactive Activities ... Sites for parents Subject: Re: Chinese remainder theorem Date: Tue, 2 Sep 1997 00:00:59 -0400 From: Alex Bogomolny Dear Tan: Yours is an example of problems solved in general case by what's known as the Chinese Remainder Theorem. You can look it up in O.Ore, "Number Theory and Its History", or H.Davenport, "The Higher Arithmetic" Both available through my bookstore. In your particular case, you are looking for a number X such that X = 1 (mod 2,3,4) and X = (mod 5) which means that divided by 2,3,4 X has the remainder 1 while divided by 5 the remainder is 0. The first three condition say that (X - 1) is divided by 2,3 and 4, i.e., by their least common multiple which is 12. Therefore, X - 1 = 12t for some integer t. From X = (mod 5) it follows that X - 1 = 4 (mod 5). Or 12t = 4(mod 5), 3t = 1 (mod 5). As you can check then, t = 5k + 2 for an integer k. Combining this with X = 12t + 1 we get X = 60k + 25. There are three numbers below 200 in this form: 25, 85 and 145. Best regards

62. Inverse Chinese Remainder Theorem - Information Technology Services In general, you can use the chinese remainder theorem. According to thechinese remainder theorem if we have moduli X,Y,Z which have no common factor, http://www.physicsforums.com/archive/t-15301_inverse_chinese_remainder_theorem.h

Technology Services Mathematics Number Theory inverse chinese remainder theorem juan avellaneda - inverse chinese remainder theorem hi all im new on the forum I wonder if is possible to find a method that proofs that a number IS NOT a solution of a set of congruences Maybe using the chinese remainder theorem best regards japam Discuss inverse chinese remainder theorem Here, Free! Muzza - inverse chinese remainder theorem Well, simply plugging in the number in the congruences and seeing it they become true should work, right? Discuss inverse chinese remainder theorem Here, Free! juan avellaneda - inverse chinese remainder theorem thats a solution but i was thinking in other possibility, a mathematical condition involving the gcd or MCM, for example im sure that it exists but i dont know how to probe that Discuss inverse chinese remainder theorem Here, Free! Hurkyl - inverse chinese remainder theorem I can't imagine any way that could be simpler than plugging your number into the equation. Is there something unusual about your problem that this is not a desirable technique? Discuss inverse chinese remainder theorem Here, Free!

63. Chinese Remainder Theorem This is the famous chinese remainder theorem. Following is my own unique methodfor solving this puzzle. Let the dividers all be prime. http://www.geocities.com/dirkie6/chinese.html

Chinese remainder theorem A certain number x is divided by 3 and a remainder of 2 results. The same number x is divided by 5 and a remainder 4 results. The same number x is divided by 7 and a remainder of 3 results. What is this number x ? This is the famous Chinese remainder theorem. Following is my own unique method for solving this puzzle. Let the dividers all be prime. Let x be the number we are looking for.So for n dividers and n remainders we have: x == a mod p (equation 1) x == a mod p (equation 2) x == a mod p (equation 3) x == a mod p (equation 4) x == a n mod p n (equation n) Now we start by multiplying equation 1 by p and equation 2 by p We then have x p == a p mod( p p ) and x p == a p mod( p p But GCD(p , p ) =1 , so we can find a r and s so that rp + sp Therefore we have xrp == ra p mod( p p ) and xsp == sa p mod( p p ) and thus xrp + xsp == ra p + sa p mod( p p So x == ra p + sa p mod( p p So x == b mod ( p p ) where b = ra p + sa p Next we look at x == b mod ( p p ) and x == a mod p We do the same as above and get x == b mod ( p p p In the end we will have x == b n mod ( p p p .....p n So x = b n will then be the number we are looking for.

64. Chinese Remainders They re called Chinese Remainder because the problem and the theorem which The chinese remainder theorem is not particularly easy to understand it http://www.delphiforfun.org/Programs/chinese_remainders.htm

Home Delphi Techniques Utilities DFF Library ... Math Topics Available Now All Programs Delphi Techniques Math Topics Utilities ... Library Units Contact Feedback Send an e-mail with your comments about this program (or anything else). Search WWW Search delphiforfun.org Problem Description What is the smallest number that can be divided by 6, leaving a remainder of 5; by 5 leaving 4; and by 4 leaving 3? Source: The Mensa (c) Puzzle Calendar for Oct 18. 2001; Here's an introduction to Chinese Remainder problems. They're called "Chinese Remainder" because the problem and the theorem which defines when they can be solved were both known to the early Chinese scholars. The earliest known example of this type of problem was published in the 3rd, 4th or 5th century AD by Chinese scholar, Sun Zi, in a book titled " Master Sun's Mathematical Manual " Here's Sun Zi's original problem: We have a number of things, but we do not know exactly how many. If we count them by threes we have two left over. If we count them by fives we have three left over. If we count them by sevens we have two left over. How many things are there?" His solution is the following poem: Three septuagenarians walking together, 'tis rare!

65. Educeth - Teaching And Learning - Informatik: Unterrichtsmaterialien - Lernumgeb chinese remainder theorem ) ist von zentraler Bedeutung in der Algebra und hatviele Anwendungen im Downloads zu Applet chinese remainder theorem http://www.educeth.ch/informatik/interaktiv/crt/

Informatik auf educeth Home Info Kontakt ... Suchen Lernumgebung zu "Chinese Remainder Theorem" KaraToJava Lernprogramme Informatik auf educeth Mathematische Themen Numerik Kleinstquadrate-Methode Numerische Verfahren der Analysis Numerische Algorithmen Werkstatt Multiplikation Diverses Roboter-Arm Fourier-Transformation Faltung Fehlerkorrigierende Codes ... Mathematik auf educeth Applet: Chinese Remainder Theorem Autor/Innen: Stefan Schmid Fachgebiet Informatik, Mathematik Schultyp Voraussetzungen Mathematische Grundkenntnisse Dauer 1-2 Lektionen Worum geht es? Der chinesische Restsatz (engl. "Chinese Remainder Theorem") ist von zentraler Bedeutung in der Algebra und hat viele Anwendungen im Bereich der Kryptographie und der Kodierungslehre. Doch was steckt eigentlich hinter diesem auf den ersten Blick kompliziert erscheinenden Theorem? Die Lernumgebung visualisiert die Kernaussage. Ausserdem kann der Benutzer das Theorem für verschiedene Zahlen verifizieren, die wesentlichen Eigenschaften erforschen oder einfach das Modulo-Rechnen und das Lösen von Kongruenzsystemen üben. Downloads zu "Applet: Chinese Remainder Theorem" Unterlagen für die Lehrperson PDF [259 KB] Word [1 MB]

66. Chinese Remainder Theorem Can Be Seen chinese remainder theorem. $ \mathop{\forall}\limits _{\substack{m, ! $ \Big(x\equiva\pmod{m} \quad \. \includegraphicswidth=8cm{chines0.eps} http://www.mtm.ufsc.br/~andsol/english/mat/china.html

See here for the ps.gz version Andrzej Solecki's cecinestpasunepeep Show proudly presents as m as n and x as x in the great Chinese Remainder Theorem index support file

67. Number Fields - Chinese Remainder Theorem chinese remainder theorem Let I 1 , . . . , I n be pairwise relatively primeideals in a ring R . Then the natural map R / n i = 1 n I i rightarrow; http://rooster.stanford.edu/~ben/maths/numberfield/crt.php

Number Fields Number Fields Sum of 2 Squares Pythagorean Triples Fermat's Last Thm ... Ramification Chinese Remainder Theorem Chinese Remainder Theorem: Let I I n be pairwise relatively prime ideals in a ring R . Then the natural map R i n I i R I R I n is an isomorphism. Proof: We first prove this for n . The result will follow by induction since I is relatively prime to I I n Assume n . Then the kernel of the map is trivial, so it remains to show that the map is surjective. In other words, for any r r R we need to find r R such that r r mod I r r mod I Find a I a I with a a , and set r a r a r My Homepage

68. 0.5.14 Chinese Remainder Theorem 0.5.14 chinese remainder theorem. 0.5 Miscellaneous Algorithms Previous0.5.13 Horner s Rule. 0.5.14 chinese remainder theorem. Scott Gasch 199907-09. http://www.darkridge.com/~jpr5/archive/alg/node139.html

Search Next: 0.5.15 Large Prime Number Up: 0.5 Miscellaneous Algorithms Previous: 0.5.13 Horner's Rule 0.5.14 Chinese Remainder Theorem Scott Gasch

69. [math/0405305] Computing Igusa Class Polynomials Via The Chinese Remainder Theor Computing Igusa class polynomials via the chinese remainder theorem. AuthorsKirsten Eisentraeger, Kristin Lauter Comments 15 pages http://arxiv.org/math.NT/0405305

Mathematics, abstract math.NT/0405305 From: Kristin Lauter [ view email ] Date: Sat, 15 May 2004 05:03:49 GMT (17kb) Computing Igusa class polynomials via the Chinese Remainder Theorem Authors: Kirsten Eisentraeger Kristin Lauter Comments: 15 pages Subj-class: Number Theory; Algebraic Geometry MSC-class: We present a new method for computing the Igusa class polynomials of a primitive quartic CM field. For a primitive quartic CM field, K, we compute the Igusa class polynomials modulo p for certain small primes p and then use the Chinese remainder theorem and a conjectural bound on the denominators to construct the class polynomials. We also provide an extension to genus 2 of Kohel's algorithm for determining endomorphism rings of elliptic curves. Our algorithm can be used to generate genus 2 curves over a finite field with a given zeta function for use in cryptography. Full-text: PostScript PDF , or Other formats References and citations for this submission: CiteBase (autonomous citation navigation and analysis) Which authors of this paper are endorsers?

70. CHINESE REMAINDER THEOREM ☼ Code Source N°29755 ☼ ☼ PHPCS.com Translate this page Beaucoup de Codes Sources pour PHP, Scripts, Script, Sources, Codes, France,Francais, French. http://www.phpcs.com/code.aspx?ID=29755

71. Chinese Remainder Problem The reason why it is called the chinese remainder Problem is because the The earliest of such works that contains the chinese remainder Problem is the http://www.math.sfu.ca/histmath/China/3rdCenturyBC/CRP1.html

Chinese Remainder Problem - The Beginning. What is it? These particular kinds of mathematical problem falls in the category of indeterminate analysis. Usually, it appears in the form as such (in modern notation): N = m1x + r1 N = m2y + r2 N = m3z + r3 Or in modern number theory notation: N r1 (mod m1) N r2 (mod m2) N r3 (mod m3) Aside: Writing N r1 (mod m1) [this means N is congruent to r1 modulo m1] means that N divided by m1 leaves r1 as the remainder. The goal here is to find the smallest positive integer satisfying the congruences states above. Origins. Now that you know what a Chinese Remainder Problem is, you must be wondering why or what has this particular kind of problem to do with Chinese Mathematical History. The reason why it is called the Chinese Remainder Problem is because the earliest versions of these congruence problems occured in early Chinese mathematical works. The earliest of such works that contains the Chinese Remainder Problem is the Sun Tzu Suan Ching (also known as Sunzi suanjing) written in approximately late third century by Sun Zi . Problem 26 (also known as the problem of Master Sun) in the third volume of the Sun Tzu Suan Ching offers the earliest recorded Chinese Remainder Problem. Problem 26 is as stated below: "We have a number of things, but we do not know exactly how many. If we count them by threes we have two left over. If we count them by fives we have three left over. If we count them by sevens we have two left over. How many things are there?" (Quoted from Sun Tze Suan Ching).

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