assig8 - particular in section 9.1, have a look at the...

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MATH 135 Algebra, Assignment 8 Due: Wed Nov 11, 8:30 am 1: Solve each of the following linear congruences. (a) 5 x 4 (mod 7) (b) 40 x 15 (mod 65) (c) 391 x 119 (mod 1003) 2: (a) Find [12] - 1 in Z 29 . (b) Solve [34] x = [18] in Z 46 . (c) In Z 20 , solve the pair of simultaneous equations [7] x + [12] y = [6] [6] x + [11] y = [13] 3: (a) Find the inverse of every invertible element in Z 15 . (b) With the help of the following list of powers of 5 mod 23, solve [11] x 18 = [15] in Z 23 . k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 5 k 1 5 2 10 4 20 8 17 16 11 9 22 18 21 13 19 3 15 6 7 12 14 4: (a) Find 17 458 (mod 13). (b) Find 47 38 54 (mod 11). (c) Find 300 k =1 k k (mod 7). 5: For this problem, you may find it useful to read some of sections 9.1 and 9.9 in the text. In
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Unformatted text preview: particular in section 9.1, have a look at the example involving long division in Z 5 on page 231, and see the Remainder Theorem 9.12 and the Factor Theorem 9.14 on page 232, and in section 9.9, look at example 9.92 on page 260. It is also worth noticing that Theorem 9.17 in section 9.1 does not always hold for polynomials over Z n . (a) Solve x 2 + 3 x + 2 0 (mod 6), then nd two dierent ways to factor the polynomial f ( x ) = x 2 + [3] x + [2] over Z 6 . (b) Solve x 2 +2 x +26 0 (mod 125), then nd two dierent ways to factor the polynomial f ( x ) = x 2 + [2] x + [26] over Z 125 ....
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This note was uploaded on 05/04/2010 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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