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).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
[Sets and number] Please use extended euclidean algorithm to find the value of a :)
Transcribed Image Text:d): 35a = 10 (45)
45 = 35x1+ lo
35=10x3 + 3
10 = 5x2 +0.
So
。 god (35₁43) = 5.
5 = 35 - 3×10 = 3§ - 3 × ( 45 - 35) = 4×351-3×45
However, 35× (-3) € 10 (mod 45)
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