hw7math109

hw7math109 - MATH 109 Assignment 7 Due: 5/23/11 #17.1...

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Unformatted text preview: MATH 109 Assignment 7 Due: 5/23/11 #17.1 Solution. By the division algorithm, we have 7684 = 4148(1) + 3536 4148 = 3536(1) + 612 3536 = 612(5) + 476 612 = 476(1) + 136 476 = 136(3) + 68 136 = 68(2) + 0 Rearranging these equations, we find m = 27 and n =- 50. 68 = 476(1)- 136(3) = 476(1)- (612(1)- 476(1))(3) = 612(- 3) + 476(4) = 612(- 3) + (3536(1)- 612(5))(4) = 3536(4) + 612(- 23) = 3536(4) + (4148(1)- 3536(1))(- 23) = 4148(- 23) + 3536(27) = 4148(- 23) + (7684(1)- 4148(1))(27) = 7684(27) + 4148(- 50) . #17.2 Solution By the division algorithm we have 7648 = 4148(1) + 3500 4148 = 3500(1) + 648 3500 = 648(5) + 260 648 = 260(2) + 128 260 = 128(2) + 4 128 = 4(32) + 0 . Rearranging these equations, we find 4 = 260- 128(2) = 260- (648- 260(2))(2) = 648(- 2) + 260(5) = 648(- 2) + (3500- 648(5))(5) = 3500(5) + 648(- 27) = 3500(5) + (4148- 3500)(- 27) = 4148(- 27) + 3500(32) = 4148(- 27) + (7648- 4148)(32) = 7648(32) + 4148(- 59) ....
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This note was uploaded on 08/02/2011 for the course MATH 109 taught by Professor Knutson during the Spring '06 term at UCSD.

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hw7math109 - MATH 109 Assignment 7 Due: 5/23/11 #17.1...

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