Unformatted text preview: x ² 1 (mod 7) x ² 4 (mod 11) An integer x satis&es the &rst equation exactly when x = 1+7 m for some integer m . Plugging this into the second equation gives 1 + 7 m ² 4 (mod 11) 7 m ² 3 (mod 11) 8 & 7 m ² 8 & 3 (mod 11) m ² 2 (mod 11) so m = 2 + 11 n for some integer n . Thus x = 1 + 7 m = 1 + 7 (2 + 11 n ) = 15 + 77 n so the solutions are the integers that are congruent to 15 modulo 17 ....
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This note was uploaded on 07/13/2011 for the course MAD 2104 taught by Professor Staff during the Spring '08 term at FAU.
 Spring '08
 STAFF

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