Sol5 - MATH 235 – Algebra 1 Solutions to Assignment 5 Solution of 1 Since 9 99 999 etc are all multiples of 9 10 n is always 1(mod 9 n ≥ 0

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Unformatted text preview: MATH 235 – Algebra 1 Solutions to Assignment 5 October 16, 2007 Solution of 1. Since 9, 99, 999, etc. are all multiples of 9, 10 n is always ≡ 1 (mod 9), n ≥ 0. So, that means that if m has digits a n ,a n- 1 ,...,a 1 ,a , m = 10 n a n +10 n- 1 a n- 1 + ... +10 a 1 + a ≡ a n + a n- 1 + ... + a 1 + a (mod 9) . This means that a number is congruent mod 9 to the sum of its digits, and if we repeatedly sum the digits until we get a single digit number, the single digit number we arrive at will be the unique number between 1 and 9 which is congruent mod 9 to our original number. Since A ≡ a (mod 9), B ≡ b (mod 9), and C ≡ c (mod 9), this means that if C = AB , c ≡ ab (mod 9), so since c and the sum of the digits of ab are both between 1 and 9, these two numbers should be equal. Solution of 2. (1) To find all solutions of the equation x 2 + x = 0 in Z / 5 Z , the easiest way is to try each element and see whether it is a solution or not. So we compute 2 + 0 = 0 , 3 2 + 3 = 12 = 2 , 1 2 + 1 = 2 , 4 2 + 4 = 20 = 0 , 2 2 + 2 = 6 = 1 , and we see that there are exactly two solutions, namely 0 and 4....
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This homework help was uploaded on 04/18/2008 for the course MATH 235 taught by Professor Goren during the Fall '07 term at McGill.

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Sol5 - MATH 235 – Algebra 1 Solutions to Assignment 5 Solution of 1 Since 9 99 999 etc are all multiples of 9 10 n is always 1(mod 9 n ≥ 0

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