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Chinese Remainder Theorem: more detail

Chinese Remainder Theorem: Applications in Computing, Coding, Cryptography by C. Ding, D. Pei, et all 1999-06

Secret Sharing Using the Chinese Remainder Theorem: Secret Sharing, Chinese Remainder Theorem, Threshold Cryptosystem, Cardinality, Access Structure, Shamir's ... Polynomial Interpolation, George Blakley

Remainder: Natural Number, Real Number Modulo Operation, Chinese Remainder Theorem, Division Algorithm, Euclidean Algorithm

A hierarchical single-key-lock access control using the Chinese remainder theorem (OSU-CS-TR) by Kim Sin Lee, 1994

Fundamental Number Theory with Applications (Discrete Mathematics and Its Applications) by Richard A. Mollin, 1998-01-31

Fundamental Number Theory with Applications, Second Edition (Discrete Mathematics and Its Applications) by Richard A. Mollin, 2008-02-21

21. Math 5410 Chinese Remainder Theorem
chinese remainder theorem. Theorem Suppose that m 1 , m 2 , , m r are pairwiserelatively prime positive integers, and let a 1 , a 2 , , a r be integers.
http://www-math.cudenver.edu/~wcherowi/courses/m5410/ctccrt.html
Chinese Remainder Theorem
Theorem : Suppose that m , m , ..., m r are pairwise relatively prime positive integers, and let a , a , ..., a r be integers. Then the system of congruences, x = a i (mod m i x m x ... x m r , which is given by:
x = a M y + a M y + ... + a r M r y r (mod M), where M i = M/m i and y i = (M i (mod m i Pf : Notice that gcd(M i , m i i all exist (and can be determined easily from the extended Euclidean Algorithm). Now, notice that since M i y i = 1 (mod m i ), we have a i M i y i = a i (mod m i i M i y i = (mod m j ) if j is not i (since m j i in this case). Thus, we see that x = a i (mod m i If there were two solutions, say x , and x , then we would have x - x = (mod m i ) for all i, so x - x = (mod M), i.e., they are the same modulo M.
Example
Find the smallest multiple of 10 which has remainder 2 when divided by 3, and remainder 3 when divided by 7. We are looking for a number which satisfies the congruences, x = 2 mod 3, x = 3 mod 7, x = mod 2 and x = mod 5. Since, 2, 3, 5 and 7 are all relatively prime in pairs, the Chinese Remainder Theorem tells us that there is a unique solution modulo 210 ( = 2x3x5x7). We calculate the M i 's and y i 's as follows:
M = 210/2 = 105; y

22. Chinese Remainder Theorem
A Mechanical Proof of the chinese remainder theorem. David M. Russinoff.This paper (ps, pdf), which was presented at ACL2 Workshop
http://www.onr.com/user/russ/david/crt.html
A Mechanical Proof of the Chinese Remainder Theorem
David M. Russinoff
This paper ( ps pdf ), which was presented at ACL2 Workshop 2000 (see slides: ps pdf ), describes an ACL2 proof of the Chinese Remainder Theorem: If m m k are pairwise relatively prime moduli and a a k are natural numbers, then there exists a natural number x that simultaneously satisfies x a i (mod m i i k The entire proof is contained in the single event file crt.lisp , except that it depends on some lemmas from the author's library of floating-point arithmetic . In order to certify this file (after obtaining and certifying the library), first replace each of the two occurrences of " /u/druss/ " with the path to the directory under which your copy of the library resides. A second event file, summary.lisp , which contains the definitions and main lemmas involved in the proof, may then be certified.

23. The Chinese Remainder Theorem
Solving the congruences x a (mod m), x b (mod n). (See descriptionof algorithm.) Enter a Enter b Enter m ( 1) Enter n( 1)
http://www.numbertheory.org/php/chinese2.html
Solving the congruences x a (mod m), x b (mod n)
(See description of algorithm Enter a:
Enter b:
Last modified 3rd September 2002
Return to main page

24. Untitled
The chinese remainder theorem and its Application in a HighSpeed RSA Crypto-Chip.Johann Großschädl IAIK, Graz University of Technology Austria.
http://www.acsac.org/2000/abstracts/48.html
16th Annual Computer Security Applications Conference
December 11-15, 2000
New Orleans, Louisiana The Chinese Remainder Theorem and its Application in a High-Speed RSA Crypto-Chip
IAIK, Graz University of Technology
Austria Read Paper (in PDF)

25. Chinese Remainder Theorem

http://www.cs.utexas.edu/users/moore/acl2/workshop-2000/final/russinoff-short/cr
This directory contains an ACL2 proof of the Chinese Remainder Theorem, as described in a paper presented at ACL2 Workshop 2000. The entire proof is contained in the single event file crt.lisp , except that it depends on some lemmas from the author's library of floating-point arithmetic . In order to certify this file (after obtaining and certifying the library), first replace each of the two occurrences of " /u/druss/ " with the path to the directory under which your copy of the library resides. A second event file, summary.lisp , which contains the definitions and main lemmas involved in the proof, may then be certified. You can download a gzipped tar file containing this file together with the two event files discussed above.

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27. The Chinese Remainder Theorem Or CRT
The chinese remainder theorem or CRT. Proposition 2.19 (Chinese RemainderTheorem for simultaneous congruences) Let be coprime. Then
http://www.maths.nott.ac.uk/personal/jec/courses/G13NUM/cnotes/node15.html
Next: The structure of Up: Congruences Previous: Some Applications Contents
The Chinese Remainder Theorem or CRT
Proposition 2.19 (Chinese Remainder Theorem for simultaneous congruences) Let be coprime. Then for every pair of integers the simultaneous congruences
have a solution which is unique modulo More generally, if then the congruences ( ) have a solution if and only if , and the solution (when it exists) is unique modulo lcm Proof . Write to satisfy the first congruence; the second then becomes or , which by Proposition 2.5 has a solution if and only if where . Uniqueness: is unique modulo , so is unique modulo To find the solution in the coprime case, write . Then we have the solution since while Example: Let . Then so the solution for general is The CRT says that there is a bijection between pairs and single residue classes when are coprime. This bijection is in fact a ring isomorphism: Theorem 2.20 (Chinese Remainder Theorem, algebraic form) Let be coprime. Then we have the isomorphism of rings Restricting to units on both sides, we have the isomorphism of groups

28. Math_class: Number Theory 101 (Chinese Remainder Theorem)
math_class Number Theory 101 (chinese remainder theorem). Disclaimersand Apologies. I said The chinese remainder theorem. I think
http://www.csh.rit.edu/~pat/math/series/nt/20020926/
: Number Theory 101 (Chinese Remainder Theorem)
I said, in the last lesson, that we would get into factoring during this lesson. I had forgotten, at the time, that I wanted to hit on the Chinese Remainder Theorem. So, factoring will have to wait until next class. In other news, I entirely blew creating a class for two weeks ago. And, last week, I was sick to the point of inertness for 80% of the week. My apologies for blowing the class two weeks ago. I hope the content is interesting enough to bring y'all back after this unscheduled hiatus.
Greatest Common Divisor (Continued)
One more interesting thing to note about the Greatest Common Divisor of two numbers (at least one of which is non-zero). The Greatest Common Divisor of two numbers is the smallest positive number which can be written as a linear combination of the two numbers. It shouldn't come as a surprise to you that the proof of this takes advantage of the Well-Ordering Principle. Just about any mathematical statement containing "smallest number" requires the Well-Ordering Principle. The proof creates a set of all positive numbers a * u + b * v where u and v are integers. It shows that the set is non-empty (one way to do this is to let

29. The Prime Glossary: Chinese Remainder Theorem
related to prime numbers. This pages contains the entry titled 'Chineseremainder theorem.' Come explore a new prime term today!
http://primes.utm.edu/glossary/page.php?next=certificate of primality

30. The Chinese Remainder Theorem
The chinese remainder theorem. The chinese remainder theorem asserts that a solutionto Sun's question exists, and the proof gives a method to find a solution.
http://modular.fas.harvard.edu/edu/Fall2001/124/lectures/lecture6/html/node2.htm
Next: Multiplicative Functions Up: Lecture 6: Congruences, Part Previous: Wilson's Theorem
The Chinese Remainder Theorem
Sun Tsu Suan-Ching (4th century AD): There are certain things whose number is unknown. Repeatedly divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the remainder is 2. What will be the number? In modern notation, Sun is asking us to solve the following system of equations:
The Chinese Remainder Theorem asserts that a solution to Sun's question exists, and the proof gives a method to find a solution. Theorem 2.1 (The Chinese Remainder Theorem) Let and such that . Then there exists such that
Proof . The equation has a solution since . Set . We next verify that is a solution to the two equations. Then and Now we can solve Sun's problem:
First, we use the theorem to find a solution to the pair of equations
Set . Step 1 is to find a solution to . A solution is . Then . Since any with is also a solution to those two equations, we can solve all three equations by finding a solution to the pair of equations
Again, we find a solution to

31. CHINESE REMAINDER THEOREM METHOD FOR FAST DECRYPTION TUTORIAL
Advanced Technology Information Processing Systems. USING THE CHINESE REMAINDERTHEOREM (CRT) FOR FAST DECRYPTION TUTORIAL. chinese remainder theorem.
http://people.atips.ca/~walus/Mont/crtexp.html
USING THE CHINESE REMAINDER THEOREM (CRT) FOR FAST DECRYPTION TUTORIAL
An important calculation in the RSA encryption scheme is the modular exponentiation M = C d (mod n). This is performed each time a part of the message is encrypted/decrypted. Both d and n are very large integers therefore this operation is very computationaly intensive. We must therefore find alternatives to the binary method for modular exponentiation. In this tutorial we will consider the use of the Chinese Remainder Theorem as a method for computing the modular exponentiation. As well we have included a interactive calculator at the end of the tutorial along with the necessary MATLAB code. The basic advantage with using the Chinese Remainder Theorem is that it allows us to split up the large modulo exponentiation into two much smaller exponentiations, one over p and one over q. These two moduli are the prime factors of n and are known. As well we can further reduce the size of the problem by using Fermat's theorem as described below. This method was first proposed by Quisquater and Couvreur.
Chinese Remainder Theorem
Definition: A residue is the remainder of a division by a number called a moduli(i.e residue of 5/7 is 2).

32. Hilbert Functions And The Chinese Remainder Theorem: Open Problems
Hilbert Functions and the chinese remainder theorem Open Problems. Notes This20 minute talk was given February 27, 1999, at the UNL Regional Workshop.
http://www.math.unl.edu/~bharbour/UNLregionwstalk.html
Hilbert Functions and the Chinese Remainder Theorem: Open Problems
Notes:
This 20 minute talk was given February 27, 1999, at the UNL Regional Workshop.
For this talk let k be any algebraically closed field.
Preliminary Problems
Problem 1 : Given points x , ... , x n of k and values v , ... , v n of k, find all polynomials f(x) in k[x] such that f(x i ) = v i for all i.
Solution : There exists a unique solution, f L (x), of degree at most n-1; it is given by the Lagrange interpolation formula:
Let g(x) = (x-x )...(x-x n
Let g i (x) = g(x)/(x-x i ) [This is a polynomial.]
Then f L (x) = (v /g (x ))g (x) + ... + (v n /g n (x n ))g n (x)
The complete set of solutions is f L (x) + (g); i.e., f L (x) + p(x)g(x), where p(x) is any polynomial. [In the usual notation, (g) is the ideal generated by g(x).]
Restatement of Problem 1 and Solution : The problem is to find all f(x) conguent mod I j to v j , for all j, where I j is the ideal (x-x j ) of the ring R=k[x]. I.e., given an element v = (v , ... , v n ) of R/I x ... x R/I n , find all elements f of R mapping to v under the homomorphism H : f -> (f , ... , f

33. CHINESE REMAINDER THEOREM
chinese remainder theorem. Let hcf (n1,n2,n3,,nr)=1. Then the system of linearcongruences xa1 mod n1. xa2 mod n2. xa3 mod n3 . xar mod nr.
http://www.bearnol.pwp.blueyonder.co.uk/Math/chinese.htm
CHINESE REMAINDER THEOREM Let hcf (n1,n2,n3,,nr)=1. Then the system of linear congruences: has a simultaneous solution, unique modulo n1n2n3nr Proof: Let n=n1n2n3nr Let X=a1N1x1 + a2N2x2 + + arNrxr so X == akNkxk == ak.1 == ak [mod nk] Suppose Y is another solution:

34. The Chinese Remainder Theorem
The chinese remainder theorem. The chinese remainder theorem statesthat if you have Q numbers N 1 to N Q which have no factors in
http://www.disappearing-inc.com/C/chineseremainder.html
Cyclopedia Cryptologia
The Chinese Remainder Theorem
The chinese remainder theorem states that if you have Q numbers N to N Q which have no factors in common, then any integer greater than or equal to and less than the product of all numbers N can be uniquely represented by a series consisting of the remainders of division by the numbers N. To walk through a simple example of how this works, let's say that N is 3 and N is 5. The Chinese remainder theorem says that every integer from to 14 will have a unique set of remainders modulo these two N's. And in fact, that's true. has a remainder of modulo 3 and a remainder of modulo 5.
1 has a remainder of 1 modulo 3 and a remainder of 1 modulo 5.
2 has a remainder of 2 modulo 3 and a remainder of 2 modulo 5.
3 has a remainder of modulo 3 and a remainder of 3 modulo 5.
4 has a remainder of 1 modulo 3 and a remainder of 4 modulo 5.
5 has a remainder of 2 modulo 3 and a remainder of modulo 5.
6 has a remainder of modulo 3 and a remainder of 1 modulo 5.
7 has a remainder of 1 modulo 3 and a remainder of 2 modulo 5.

35. Chinese Remainder Theorem
chinese remainder theorem. x1 = x % z1, x = x1 + j*z1 for some integerj. x2 = x % z2, x = x2 + k*z2 for some integer k. Let inv(y,n
http://www.csc.gatech.edu/~copeland/6086/q_and_a/990714_Chinese_RT.html
Chinese Remainder Theorem x1 = x % z1, x = x1 + j*z1 for some integer j x2 = x % z2, x = x2 + k*z2 for some integer k Let inv(y,n) denote the multiplicative inverse of y modulus n. This exists if y and n are "relatively prime" (have no common factors). The equal sign underlined, , indicates modular "equivalence." Given that z1 and z2 are relatively prime, then inv(z1,z2) and inv(z2,z1) both exist. Also by Euclid's Theorem (since the gcd of z1 and z2 is 1), we can find integers u and v such that: u*z1 + v*z2 =1 1 mod z2 1 mod z1 multiply above by x and exchange sides to get: x = x*u*z1 + x*v*z2 =1 Since (k*z2 u*z1) % (z1z2) = x x2*z1*inv(z1,z2) + x1*z2*inv(z2,z1) mod z1z2 Suppose z1, z2, and z3 are relatively prime: then z1 is relatively prime to z2z3, and we can find x23 = x % (z2z3), inv(z1, z2z3), and inv(z2z3, z1) x x23*z1*inv(z1, z2z3) + x1*z2*z3*inv(z2z3, z1)

36. Chapter 7: The Chinese Remainder Theorem
Chapter 7. The chinese remainder theorem. 7.1 Solving Two Congruences 7.2 AMore General Theorem 7.3 Solving Lots of Congruences 7.4 Explicit Formulas
http://www.mathlab.mtu.edu/numbertheory/web/chinese.html
Chapter 7
The Chinese Remainder Theorem
Solving Two Congruences
A More General Theorem
Solving Lots of Congruences
Explicit Formulas ...
DNT Table of Contents

37. Chinese Remainder Theorem
Rings, chinese remainder theorem. Search Site map Contact us Join ourmailing list Books Back to Theory. chinese remainder theorem. The
http://www.mathreference.com/ring,chr.html
Rings, Chinese Remainder Theorem
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Rings
Use the arrows at the bottom to step through Ring Theory.
Chinese Remainder Theorem
The chinese remainder theorem was developed for modular arithmetic , but it generalizes to ideals in a commutative ring r. Let h h ... h n be a set of coprime ideals. By coprime, we mean the sum of any two ideals spans the ring. Let j be the product of all these ideals. We will prove r/j is isomorphic to the direct product of the quotient rings r/h i , as i runs from 1 to n. An element in r/j can be mapped to the ith component in the direct product via r/h i . This is a well defined ring homomorphism, since each h i wholly contains j. We need to show it is 1-1 and onto. Focus on h . We know x + y = 1 for some x in h and y in h . Do the same for each h i in the set. Multiply all these equations together, and something in h + something in the product of the other ideals gives 1. Write this as x + y = 1. Reduce mod h , and y = -1. If y

38. Chinese Remainder Theorem
Modular Mathematics, chinese remainder theorem. Search Site map Contactus Join our mailing list Books chinese remainder theorem.
http://www.mathreference.com/num-mod,chr.html
Modular Mathematics, Chinese Remainder Theorem
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Numbers
Modular Mathematics
Use the arrows at the bottom to step through Modular Math.
Chinese Remainder Theorem
Given a set of values c c ... c n , and a set of mutually coprime moduli m m ... m n , is there an integer x such that x = c i mod m i for each i in 1 through n? Let z be the product of all the moduli. If x is a solution then so is x plus any multiple of z. If w is not a multiple of z, say w is not divisible by m , then x+w will not equal c mod m , and x+w will not be a solution. The solution, if it exists, is well defined mod z. To show that a solution exists, we simply construct one. Let a i be the product of all the moduli other than m i . Verify that a i and m i are coprime. Let b i be the inverse of a i mod m i . Finally, let x be the sum of a i b i c i for all i in 1 ... n. Verify that x satisfies all n equations simultaneously. If the original moduli are not coprime, split each equation up into a set of equations by factoring the composite modulus into prime powers. Then consider all the equations together. Equations sharing a common prime modulus are either redundant or inconsistent. An inconsistent example is x = 4 mod 6 and x = 11 mod 15 . This would force x = 1 mod 3 and x = 2 mod 3, which is impossible. The example x = 4 mod 6 and x = 7 mod 15 has the solution x = 22 mod 30.

39. Modified Chinese Remainder Theorem And Its Application To Proxy Signatures
Processing September 21 24, 1999 Wakamatsu, Japan, p. 146 Modified ChineseRemainder Theorem and Its Application to Proxy Signatures. PDF.
http://www.computer.org/proceedings/icpp/0353/03530146abs.htm
1999 International Workshops on Parallel Processing September 21 - 24, 1999 Wakamatsu, Japan p. 146 Modified Chinese Remainder Theorem and Its Application to Proxy Signatures Chuan-Kun Wu, Vijay Varadharajan University of Western Sydney Chinese Remainder Theorem has been used for hundreds of years and has been applied to many domains such as integers and polynomials. An assumption made is that the component moduli are pairwise co-prime. In this paper, first we remove this assumption; then we give an algorithm to find whether a given system of congruent equations has a solution, and if so, how to find the solution in an efficient manner. Further we apply the modified Chinese Remainder Theorem to design proxy signatures. The full text of icpp is available to members of the IEEE Computer Society who have an online subscription and an web account

40. Chinese Remainder Theorem Corollary
chinese remainder theorem Corollary. I don't have the proof here, but I'll getaround to writing it up. Anyway, the chinese remainder theorem is Theorem.
http://begghilos2.ath.cx/~jyseto/Academia/CRTC.html
About QYV ] [Academia] [ Battletech Homepages Journal Links ... Site Map
Chinese Remainder Theorem Corollary
Some Russian guy (if he wasn't Russian, then I'm Stalin!) and I came up with this corollary during class in first year (Math 135). I ended up proving this in the exam and using it to make some of the questions trivial to solve. I don't have the proof here, but I'll get around to writing it up. Anyway, the Chinese Remainder Theorem is:
Theorem
If GCD(m , m ) = 1 then, for any choice of integers a and a , the simultaneous congruences x == a (mod m ) and x == a (mod m have a solution. Moreover, if x = x is one solution, the complete solution is x == x (mod m m
Corrollary
If GCD(m , m ) = 1 then there exists two integers a and a , where (m ) x == (a m m ) (mod m m More general case If GCD(m i , m j ) = 1 then, for any choice of integers a , a , ... , a n the simultaneous congruences x == a (mod m
x == a (mod m
x == a n (mod m n have a solution. Moreover, if x = x is one solution, the complete solution is x == x (mod m m ... m n Therefore the corrollary is (m n ) x == (a P m / m P m / m n P m / m n ) (mod P m Where P m = m m ... m

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