Determine all heltallösningar the Diophantine equation such that both x and y is between 100 and 380 (including 100 and 380).
i wanna solve this problem with Euclidean algorithm
I don't know whether this is the method you're supposed to use, but it works.
You have 5 - 4 = 1 , so, 1000*5 - 1000*4 = 1000. .................... (1)
Also , 4*5 - 5*4 = 0, so 4n*5 - 5n*4 = 0 for any n, .................. (2)
Subract (2) from (1) and you have
(1000 - 4n)*5 - (1000 - 5n)*4 = 1000.
or,
(5n - 1000)*4 - (4n - 1000)*5 = 1000.
Now work out what value(s) you need for n.
Where are you confused? Bob has given you a way to find the solution! If you don't tell us what you don't understand it is impossible to help you.
Facts you should know from lecture or your text book are:
if
The Ecliddean alogrithm can find the GCD of a and b and express them as
(1)
If the GCD does not divide c the equation has no integer solutions if it does then there exists an integer q such that
If we multiply equation (1) by q this gives
Now
and are a solution of the problem.
All other solutions can be found using the formula (this should also be in your text book or notes)
and
Where m is some integer.
P.S this is exactly the solution Bob gave you.
Try to work these steps, and if you get stuck post back with your work so we can see what you have tried.
idk but im familiar with this thing... this is how i use to solve it and teacher solved problem like this.... idk but im really confused what i shall do next on that problem
17x+6y=250
particulate solve:
17x+6y=0
17=2*6+5
6=1*5+1
(backwards)
5=17-2*6
1=6-5
(now we use this backwards)
1=6-5 (we can rewrite 5 with 5=17-2*6)
1=6-(17-2*6)
1=3*6-1*17 (just gather all 6's)
homogeneous solve
17x+6y=250
x=-250+6k (remember the y=-1)
y=750-17k (remember x=3)
Hello, Petrus!
Here is a very primitive solution . . .
Determine all solutions the Diophantine equation:
such that both and are between 100 and 380 inclusive.
I want to solve this problem with the Euclidean algorithm.
Solve for .[1]
Since is an integer, must be a multiple of 5: .
Then [1] becomes: .
Now we have:
. . . . . . . . . .
We also have:
. . . . . . . . . . . .
Hence: .