Working Backwards from the GCD: Finding the Solution to a GCD Calculation

  • Thread starter Thread starter Math100
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
5 replies · 2K views
Math100
Messages
823
Reaction score
234
Homework Statement
Find a solution of 621m+483n=k, where k is the gcd of 621 and 483.
Relevant Equations
None.
First, we start to calculate the gcd(621, 483).
Applying the Euclidean algorithm produces:
621=1*483+138
483=3*138+69
138=2*69.
Thus gcd(621, 483)=69.

And now I'm stuck, because I don't know how to find the solution of this after finding out the gcd of 621 and 483. I was told to work/calculate backwards starting from 69, for example, like this:

69=483-(3*138)
?=?
 
Physics news on Phys.org
Start at the end:
\begin{align*}
138=2\cdot 69 & \Longrightarrow 483=3\cdot 138+69 =3\cdot(2\cdot 69)+69=7\cdot 69 \\
483=7\cdot 69 & \Longrightarrow 621=1\cdot 483 +138 = \ldots
\end{align*}
 
Last edited:
Reply
  • Like
Likes   Reactions: Math100
fresh_42 said:
Start at the end:
$$
138=2\cdot 69 \Longrightarrow 483=3\cdot 138+69 =3\cdot(2\cdot 69)+69=7\cdot 69
483=7\cdot 69 \Longrightarrow 621=1\cdot 483 +138 = \ldots
$$
Sorry, I don't understand this, as this was written in latex.
 
Math100 said:
Sorry, I don't understand this, as this was written in latex.
Now it renders. Refresh the page (F5).
 
Reply
  • Like
Likes   Reactions: Math100
At the end, you can use ##4\cdot 7 - 3\cdot 9 =1## and multiply it with ##69.##
 
Reply
  • Like
Likes   Reactions: Math100 and hutchphd