210hw6solns - Math 210 Homework 6 SOLUTIONS (possibly a...

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Math 210 Homework 6 SOLUTIONS (possibly a little terse in some cases) 1. Let a and b be integers. Prove that 3 a + b is divisible by 7 if and only if a + 5 b is divisible by 7. SOLUTION. Suppose 7 | (3 a + b ). Note that 7 | 14 b . Therefore, 7 | ± (3 a + b ) + 14 b ² = 3 a + 15 b = 3( a + 5 b ). But 7 and 3 are relatively prime. So 7 | ( a + 5 b ). Similarly, suppose 7 | ( a + 5 b ). Note that 7 | 14 a . Therefore, 7 | ± ( a + 5 b ) + 14 a ² = 15 a +5 b = 5(3 a + b ). But 7 and 5 are relatively prime. So 7 | (3 a + b ). 2. True or false? In each case, if true, give a general proof. If false, give an explicit counterexample. (All variables represent integers.) (i) If a | b and a | c , then a | bc . SOLUTION. True. If a | b then b = ax . If a | c then c = ay . Then bc = axay , which is divisible by a . (ii) If a | c and b | c , then ab | c . SOLUTION. False. For example, let a = 2, b = 6, and c = 30. Then a | c and b | c , but ab = 12 does not divide c . (iii) If a | b and c | d , then ac | bd . SOLUTION.
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210hw6solns - Math 210 Homework 6 SOLUTIONS (possibly a...

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