210hw6solns

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online