M135F08A7S

M135F08A7S - MATH 135 Fall 2008 Assignment #7 Solutions...

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MATH 135 Fall 2008 Assignment #7 Solutions Hand-In Problems 1. Considering a modulo 11, we obtain a 10 n r n + 10 n - 1 r n - 1 + ··· + 10 r 1 + r 0 (mod 11) ( - 1) n r n + ( - 1) n - 1 r n - 1 + ··· + ( - 1) r 1 + r 0 (mod 11) (since 10 ≡ - 1 (mod 11)) r 0 - r 1 + ··· + ( - 1) n - 1 + ( - 1) n (mod 11) Therefore, a r 0 - r 1 + ··· + ( - 1) n - 1 + ( - 1) n (mod 11), and so 11 | a if and only if a 0 (mod 11) if and only if r 0 - r 1 + ··· + ( - 1) n - 1 + ( - 1) n 0 (mod 11) if and only if 11 | r 0 - r 1 + ··· + ( - 1) n - 1 + ( - 1) n , as required. 2. For 23 b 2 a 4 to be divisible by 72, it must be divisible by 8 and by 9. By (a), 23 b 2 a 4 is divisible by 8 if and only if 2 a 4 is divisible by 8. Checking the possibilities, 2 a 4 is divisible by 8 only when a = 2 or a = 6. Thus, 23 b 2 a 4 = 23 b 224 or 23 b 264. For 23 b 224 to be divisible by 9, we need 2 + 3 + b + 2 + 2 + 4 0 (mod 9) or b ≡ - 13 (mod 9) or b = 5 since b is a digit. For 23 b 264 to be divisible by 9, we need 2 + 3 + b + 2 + 6 + 4 0 (mod 9) or b = - 17 (mod 9) or b = 1 since b is a digit. Thus, for 23 b 2 a 4 to be divisible by 72, we must have ( a, b ) = (2 , 5) or (6 , 1). (We should check that each of these possibilities is indeed divisible by 72. Note that 235 224 ÷ 72 = 3267 and 231 264 ÷ 72 = 3212 so both are indeed divisible by 72.) 3. (a) We convert to the linear Diophantine equation 1713 x + 2000 y = 851 and solve using the EEA: 0 1 2000 1 0 1713 - 1 1 287 1 6 - 5 278 5 - 7 6 9 1 216 - 185 8 30 - 223 191 1 1 2000 - 1713 0 8 Therefore, gcd (1713 , 2000) = 1, so there is a unique solution mod 2000. Since 1713( - 223) + 2000(191) = 1, then, multiplying by 851, we obtain 1713( - 189773) + 2000(162541) = 851, so the solution to the congruence is x ≡ - 189773 227 (mod 2000). (b) Since 2 | 1426 and 2 | 2000, then 2 | gcd(1426 , 2000). (In fact, gcd(1426 , 2000) = 2.) Since 2 6 | 851, then there is no solution.
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(c) We make a table modulo 12 to solve x 2 - 2 x 0 (mod 12): Modulo 12 x 0 1 2 3 4 5 6 7 8 9 10 11 2 x 0 2 4 6 8 10 0 2 4 6 8 10 x 2 0 1 4 9 4 1 0 1 4 9 4 1 x 2 - 2 x 0 11 0 3 8 3 0 11 0 3 8 3 so x 0 , 2 , 6 , 8 (mod 12). (d) Solution 1 Since 8 x 12 (mod 52) and each of 8, 12 and 52 is divisible by 4, then 2 x 3 (mod 13). This also means that 2 x 16 (mod 13). Since gcd(2 , 13) = 1, we can divide out the common factor of 2 to obtain x 8 (mod 13). Solution 2 Since gcd (18 , 52) = 4 and 4 | 12, then there are 4 solutions modulo 52 (or 1 solution modulo 13). By inspection, x = 8 is a solution, so the complete solution is x 8 , 8 + 52 4 , 8 + 2 ( 52 4 ) , 8 + 3 ( 52 4 ) (mod 52) 8 , 21 , 34 , 47 (mod 52) or x 8 (mod 13) . 4. (a) We want to solve [23][
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This note was uploaded on 02/23/2009 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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M135F08A7S - MATH 135 Fall 2008 Assignment #7 Solutions...

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