Solving Linear Diophantine Equation with the Euclidean Algorithm
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- Опубліковано 14 чер 2024
- Solving Linear Diophantine Equation with the Euclidean Algorithm
Explore the fascinating world of solving linear Diophantine equations using the powerful Euclidean Algorithm! Join us on a journey through the step-by-step process of finding integer solutions to these intriguing mathematical puzzles. Unlock the secrets of Diophantine equations and elevate your problem-solving skills with this comprehensive tutorial. #lineardiophantineequation #euclideanalgorithm #mathematics #mathtutorial #problemsolving #algebra
Topics Covered:
1. Understanding the basics of Linear Diophantine equation in integers.
2. Analyzing the unique method of Euclidean Algorithm.
3. Step-by-step approach to solving equation for integers.
4. Tips and tricks for solving linear Diophantine equations
5. Particular and general solutions.
Timestamps:
0:00 Introduction
0:45 Steps of linear Diophantine equation
1:48 GCD
3:02 Euclidean algorithm
6:44 Particular solution
8:20 General solutions
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Thanks for Watching!
@infyGyan
Super wonderful explanation...thanks for sharing
Thanks for watching 🙏
(x,y) =(3,-7),(12,-29),(21,-51),(30,-73),(39,-95),(48,-117)= any combo like x number(first number of x is 3 then always add 9) and for value of( y is then first number is -7 then always add -22)
x=(9n-6) and y=(15-22n) where n is the number of terms
👍👍👍
66x + 27y = 9
( x , y ) = ( - 6 + 9k , 15 - 22k )
= ( - 6 , 15 ) + k ( 9 , - 22 )
k € Z
😊🤪👋
But why haven't you divided the whole equation by gcd(66, 27) = 3 right off at the very beginning of the solution? That would have given us a much simpler (but equivalent) form of the original equation: *22x + 9y = 3,* and we would have gotten exactly the same set of solutions anyway.
But the presented solution is absolutely correct up to every tiny detail, of course. 👍
Yes we can take that way too with gcd 1.
Thanks 👍
another approach
66x + 27y = 9 => 22x + 9y = 3 => x ≡ 0 (mod 3)
let x = 3k, 66k + 9y = 3 => 22k + 3y = 1 => k ≡ 1 (mod 3)
let k = 3n +1, 22(3n +1) + 3y = 1, 22n + y + 7 = 0 => y = -7 - 22n
22*3k*(3n + 1) + 3*(-7 - 22n ) = 1
66kn + 22k - 21 - 66n = 1 => 3kn + k - 3n - 1 = (k - 1)*(3n + 1) = 0
so k = 1, x = 3k*(3n + 1) = 3 + 9n, y = -7 - 22n
(x,y) =(3 + 9n, -7 - 22n) which is same as (-6 +9k, 15 - 22k) with n = k -1
Thanks for sharing!!