# H4 - 1 mod 8 The solutions of the ﬁrst congruence are numbers of the form 2 5 k for some integer k Substitute into the second congruence 3(2 5 k

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1 Homework IV: June 30, 2009 1 . 79 . Let m be a positive integer, and let m 0 be the integer obtained from m by reversing its (decimal) digits. Prove that m - m 0 is a multiple of 9. Let a = d k ··· d 1 d 0 , where the d i are the decimal digits of a positive integer a . Let r ( m ) be the remainder after dividing d k + ··· + d 1 + d 0 by 9, By Example 1.66, r ( m ) is equal to the remainder after dividing m by 9. Since m 0 is obtained from m by reversing its digits, the family of digits of m (with possible repetitions) is the same as the family for m 0 . Therefore, r ( m 0 ) = r ( m ). But r ( m - m 0 ) r ( m ) - r ( m 0 ) mod 9 (by Example 1.66), and so r ( m - m 0 ) = 0; that is, 9 | ( m - m 0 ). 1 . 88 . Prove that if p is a prime and a 2 1 mod p , then a ≡ ± 1 mod p . Now a 2 1 mod p means that p | a 2 - 1. Of course, a 2 - 1 = ( a + 1)( a - 1). Since p is prime, Euclid’s Lemma applies to give p | ( a + 1) or p | ( a - 1); that is, a ≡ - 1 mod p or a 1 mod p . 1 . 91 ( i ) . Find all the solutions of the system x 2 mod 5 3 x
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Unformatted text preview: 1 mod 8 . The solutions of the ﬁrst congruence are numbers of the form 2 + 5 k , for some integer k . Substitute into the second congruence: 3(2 + 5 k ) ≡ 1 mod 8 6 + 15 k ≡ 1 mod 8 15 k ≡ -5 mod 8 . We can simplify this by replacing any number by its remainder mod 8 (but we don’t have to simplify): 7 k ≡ -5 mod 8 . (We could also have replaced-5 by 3.) Since 7 · 7 = 49 ≡ 1 mod 8, we have k ≡ -35 mod 8. Therefore, x = 2 + 5 k = 2 + 5(-35) =-173 . Therefore, the simultaneous solutions are all numbers of the form-173 + 40 ` . (Of course, we could replace-173 by its remainder mod 40, so all numbers of the form 27 + 40 ` ; these two sets of numbers are the same.)...
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## This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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