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# sol5 - MATH 235 Algebra 1 Solutions to Assignment 5...

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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 0 , m = 10 n a n +10 n - 1 a n - 1 + . . . +10 a 1 + a 0 a n + a n - 1 + . . . + a 1 + a 0 (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 0 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. (2) Here we compute in Z / 6 Z 0 2 + 0 = 0 , 3 2 + 3 = 12 = 0 , 1 2 + 1 = 2 , 4 2 + 4 = 20 = 2 , 2 2 + 2 = 6 = 0 , 5 2 + 5 = 30 = 0 , hence this time we find 4 solutions, namely 0, 2, 3, and 5.

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