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Unformatted text preview: 1 mod 8 . The solutions of the ﬁrst congruence are numbers of the form 2 + 5 k , for some integer k . Substitute into the second congruence: 3(2 + 5 k ) ≡ 1 mod 8 6 + 15 k ≡ 1 mod 8 15 k ≡ 5 mod 8 . We can simplify this by replacing any number by its remainder mod 8 (but we don’t have to simplify): 7 k ≡ 5 mod 8 . (We could also have replaced5 by 3.) Since 7 · 7 = 49 ≡ 1 mod 8, we have k ≡ 35 mod 8. Therefore, x = 2 + 5 k = 2 + 5(35) =173 . Therefore, the simultaneous solutions are all numbers of the form173 + 40 ` . (Of course, we could replace173 by its remainder mod 40, so all numbers of the form 27 + 40 ` ; these two sets of numbers are the same.)...
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Algebra

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