Disc-a10

Disc-a10 - x ² 1(mod 7 x ² 4(mod 11 An integer x...

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1. Find all solutions ( in standard form ) to the equation 6 x = 9 in Z 15 . 6 0 = 0 6 5 = 0 6 10 = 0 6 1 = 6 6 6 = 6 6 11 = 6 6 2 = 12 6 7 = 12 6 12 = 12 6 3 = 3 6 8 = 3 6 13 = 3 6 4 = 9 6 9 = 9 6 14 = 9 so the solutions are 4 , 9 , and 14 . 2. Find all solutions ( in standard form ) to the equation x 2 = 1 in Z 8 . 0 2 = 0 1 2 = 1 2 2 = 4 3 2 = 1 4 2 = 0 5 2 = 1 6 2 = 4 7 2 = 1 so the solutions are 1 , 3 , 5 , and 7 . 3. Find a solution ( in standard form ) to the equation 37 x = 1 in Z 1001 . x and y such that 37 x + 1001 y = 1 . 37 x + 1001 y x y q 1001 0 1 37 1 0 27 2 ± 27 1 18 1 487 ± 18 So 37 487 ± 1001 18 = 1 which means that x = 487 is a solution to the equation 37 x = 1 in Z 1001 . 4. Find all integer solutions to the pair of equations
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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.

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