CSE20 Lecture 10

# CSE20 Lecture 10 - 1 CSE 20 Discrete Math Review before...

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Unformatted text preview: 1 CSE 20 Discrete Math Review before mid-term 2 CK Cheng 2.17.2010 2 Residual Number System Residual Number System: Show the operation of 19 x 15 in a residual number system with moduli ( m 1 , m 2 , m 3 ) = (5, 13, 14). Lets start by showing some basic work and expanding the problem: 19 ( x 1 , x 2 , x 3 ) = (4, 6, 5) x 15 ( y 1 , y 2 , y 3 ) = x (0, 2, 1) 285( R 1 , R 2 , R 3 ) (0, 12, 5) Just as we did for the multiplication of 19 and 15, we multiply each item of the top row with the corresponding value in the bottom row. Staring with 4 times 0, then 6 times 2, then 5 times 1. 3 The Chinese Remainder Theorem states: ( ) % M N i = 1 Z= . M i S i R i . . ( ) % M 3 i = 1 19 15 = . M i S i R i . . Applying it to our problem we get 4 There are three things we really need to find: M i , S i , and R i . Lets start with M i . Also note: (to find big M) M = m1 x m2 x m3 M = 5 x 13 x 14 M 1 = m2 x m3 = 13 x 14 = 182 M 2 = m3 x m1 = 14 x 5 = 70 M 3 = m1 x m2 = 5 x 13 = 65 5...
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## CSE20 Lecture 10 - 1 CSE 20 Discrete Math Review before...

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