d): 35a = 10 (45) 45 = 35x 1 + lo 35 = 10 x 3 + (3) 10 = 5x2 +0. so gcd (35.45) = 5, 5 = 35- 3×10 = 35 - 3× (45 -35) = 4×351-3×45 However, 35× (-3) € 10 (mod 45).
d): 35a = 10 (45) 45 = 35x 1 + lo 35 = 10 x 3 + (3) 10 = 5x2 +0. so gcd (35.45) = 5, 5 = 35- 3×10 = 35 - 3× (45 -35) = 4×351-3×45 However, 35× (-3) € 10 (mod 45).
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.9: Dividing Real Numbers
Problem 10MRE
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