Chinese Reminder Theorem PDF The Chinese Remainder Theorem enables

Math 127: Chinese Remainder Theorem

Example 2. Find x such that 3x ? 6 (mod 12). Solution. Uh oh. This time we don't have a multiplicative inverse to

Chinese Reminder Theorem

The Chinese Remainder Theorem enables one to solve simultaneous equations with respect For example in the first equation for y1

The Chinese Remainder Theorem

07-Jun-2014 The basic form is about a number n that divided by some divisors and leaves remainders. Page 4. Title. Definition. Example. Principle. More ...

General Secret Sharing Based on the Chinese Remainder Theorem

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Compartmented Secret Sharing Based on the Chinese Remainder

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Remainder Theorem Definition And Example

Remainder and Factor Theorems Precalculus Socratic. The Chinese Remainder. Theorem NRICH Millennium. In examples of polynomials determined by its factor

The Chinese Remainder Theorem Theorem. Let m and n be two

Example: Solve the system of congruences x ? 1 (mod 7) x ? 3 (mod 10). Note that the hypotheses of the Chinese re- mainder theorem are satisfied in this

Large Numbers the Chinese Remainder Theorem

https://www.math.tamu.edu/~stephen.fulling/chinese.pdf

Chinese Remainder Theorem

Example.Find a solution to

x88 (mod 6) x100 (mod 15) Solution 1:From the rst equation we know we wantx88 = 6kfor some integerk, so xis of the formx= 88 + 6k. So from the second equation, we also want 88 + 6k100 (mod 15), so we want 6k12 (mod 15). Use the extended Euclidean Algorithm to nd that 15(1)+6(2) = 3. Multiply through by 4 to get 15(4)+6(8) = 12. Thus 6(8)12 (mod 15), and we can takek=8. This results inx= 88 + 6(8) = 40.

Solution 2:Simplify the two equations rst, to:

x4 (mod 6) x10 (mod 15) From the rst equation we know we wantx4 = 6kfor some integerk, soxis of the form x= 4 + 6k. So from the second equation, we also want 4 + 6k10 (mod 15), so we want

6k6 (mod 15). Use the extended Euclidean Algorithm to nd that 15(1) + 6(2) = 3.

Multiply through by 2 to get 15(2) + 6(4) = 12. Thus 6(4)6 (mod 15), and we can takek=4. This results inx= 4 + 6(4) =20.

Solution 3:Simplify the two equations rst, to: