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M135F08A6S

# M135F08A6S - MATH 135 Fall 2008 Assignment#6 Solutions...

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MATH 135 Fall 2008 Assignment #6 Solutions Hand-In Problems 1. (a) Answer : YES Since ( - 14) - 22 = - 36 = ( - 2)(18), then 18 | ( - 14) - 22, so - 14 22 (mod 18). (b) Answer : YES Since 2473 - ( - 26) = 2499 = 147(17), then 17 | 2473 - ( - 26), so 2473 ≡ - 26 (mod 17). (c) Answer : YES Note that 8 2 64 4 (mod 12) and 8 3 8 2 8 4(8) 32 8 (mod 12). Thus, the powers of 8 alternate being congruent to 8 and 4 modulo 12, with all even powers being congruent to 4 modulo 12. (d) Answer : 502 n 0 (mod 4) if and only if 4 | n . Since 2008 = 4(502), then there are 502 multiples of 4 in this range. (e) Answer : 2 Since 8 ≡ - 1 (mod 3), then 8 243 ( - 1) 243 ≡ - 1 (mod 3). Since the remainder is positive, then we rewrite this as 8 243 ≡ - 1 2 (mod 3). Thus, 8 243 and 2 have the same remainder when divided by 3, which is 2. (f) Answer : NO Since 8 1 (mod 7) and 13 ≡ - 1 (mod 7), then 8 24 + 13 12 1 24 + ( - 1) 12 1 + 1 2 (mod 7) In other words, 8 24 + 13 12 is not congruent to 0 modulo 7, so is not divisible by 7. (g) Answer : 0 We look at small powers of 3 and of 5, until we find one congruent to 1 or - 1 modulo 13: 3 2 9 (mod 13) 3 3 27 1 (mod 13) and 5 2 25 ≡ - 1 (mod 13) Thus, 3 47 = (3 3 ) 15 3 2 1 15 3 2 9 (mod 13). Also, 5 74 = (5 2 ) 37 ( - 1) 37 ≡ - 1 (mod 13). Next, we note that - 9 4 (mod 13), so we look at small powers of 4 modulo 13: 4 2 16 3 (mod 13) 4 3 64 ≡ - 1 (mod 13) Thus, ( - 9) 10 4 10 (4 3 ) 3 4 1 ( - 1) 3 4 ≡ - 4 (mod 13). Therefore, 3 47 5 74 + ( - 9) 10 (9)( - 1) - 4 ≡ - 13 0 (mod 13) Therefore, the remainder is 0. (h) Answer : 11 For 14 ≡ - 58 (mod m ), we need m | 14 - ( - 58) = 72, so m is a positive integer divisor of 72 which is larger than 1. Since 72 = 2 3 3 2 , then 72 has (3 + 1)(2 + 1) = 12 positive divisors, one of which is 1. Thus, 72 has 11 positive divisors larger than 1.

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2. We prove the result by induction on n . Base Case Suppose n = 1. Since a b (mod m ), then a 1 b 1 (mod m ), as required. Induction Hypothesis Suppose that the result is true for n = k , for some k P . That is, suppose that a k b k (mod m ). Induction Conclusion Consider n = k + 1. We know that a b (mod m ) and a k b k (mod m ) by the Induction Hypothesis.
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