# 1153 - Section 2.3 3. The modulus 4, 3, 7 are pairwise...

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Section 2.3 3. The modulus 4 , 3 , 7 are pairwise relatively prime, so we can apply the algo- rithm of the Chinese Remainder Theorem: we ﬁrst solve: 21 c 1 1 mod 4 28 c 2 1 mod 3 12 c 3 1 mod 7 One may take: c 1 = 1 c 2 = 1 c 3 = 3 From this, one obtains the solution to the simultaneous congruence equation: x 1 mod 4 x 0 mod 3 x 5 mod 7 by the formula x 1 × 21 × 1 + 0 × 28 × 1 + 5 × 12 × 3 mod 84 33 mod 84 7. One solves the congruence equation: 5 x 1 mod 6 (1) and obtains x ≡ - 1 mod 6 Similarly, one solves: 4 x 13 mod 15 (2) and obtains (noting that 4 × 4 1 mod 15) x 4 × 13 7 mod 15 So we have to solve the simultaneous congruences x ≡ - 1 mod 6 (3) x 7 mod 15 (4) In this case, the modulus 6 and 15 are not relatively prime, so we cannot apply the Chinese remainder theorem. However, from (7), we may write x = - 1+6 k , and substitute into (8), we obtain the congruence equation: - 1 + 6 k 7 mod 15 1

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or 6 k 8 mod 15
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## This note was uploaded on 01/09/2010 for the course MATH 115 taught by Professor Mok during the Fall '07 term at University of California, Berkeley.

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1153 - Section 2.3 3. The modulus 4, 3, 7 are pairwise...

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