Review3-2011-Solns

Review3-2011-Solns - MATH 127: Exam 3 Review Monday, April...

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Unformatted text preview: MATH 127: Exam 3 Review Monday, April 18, 2011 1. If x Z is odd then x 1 , 3 , 5 or 7 (mod 8). Squaring both sides we see x 1 , 1 , 1 or 1 (mod 8) or simply x 1 (mod 8). Thus x 2 = 8 k + 1. 2. ( ca,cb ) = cam + cbn where this is the smallest positive integer that can be written as a linear combination of ca and cb . cam + cbn = | c || am + bn | where | am + bn | is the smallest positive integer that can be written as a linear combination of a and b . Thus ( ca,cb ) = | c || am + bn | = | c | ( a,b ) as desired. 3. If d | (3 k + 2) and d | (5 k + 3) then d | ((3 k + 2)(5) + (5 k + 3)(- 3)) or simply d | 1. Thus (3 k + 2 , 5 k + 3) | 1 and hence (3 k + 2 , 5 k + 3) = 1 4. Euclidean Algorithm: 1776 = 1492(1) + 284 1492 = 284(5) + 72 284 = 72(3) + 68 72 = 68(1) + 4 68 = 4(17) + 0 Thus (1776 , 1492) = 4. Working backwards we are able to write 4 as a linear combination of 1776 and 1492 by 4 = 1492(25) + 1776(- 21) 5. Let a = p a 1 1 p a k k and b = p b 1 1 p b k k where a i ,b i 0 for all i . a 3 | b 2 implies p 2 b 1 1 p 2 b k k = p 3 a 1 + c 1 1 p 3 a k + c k k where each c i 0. Thus 3 a i 2 b i for all i . Hence a i 2 / 3 b i b i for all i . Hence a | b . 6. (a) Suppose that p a || m and p b || n . Then m = p a Q and n = p b R where both Q and R are products of primes other than p . Hence mn = ( p a Q )( p b R ) = p a + b QR . It follows that p a + b || mn since p does not divide QR . (b) Suppose that p a || m and p b || n with a 6 = b . Then m = p a Q and n = p b R where both Q and R are products of primes other than p . Suppose, WLOG, that a = min ( a,b ). Then m + n = p a Q + p b R = p min ( a,b ) ( Q + p b- a R ). Then p 6 | ( Q + p b- a R ) because...
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Review3-2011-Solns - MATH 127: Exam 3 Review Monday, April...

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