Question

Write gcd (210, 45) as a linear combination of 210 and 45 then use it to solve Diophantine 210x +45y = 60
Please, help

Answer 1

What does Diophantine mean?

Answer 2

My question is on second part.
gcd (210, 45) =15
210 = 45(4) + 30 (a)
45= 30 + 15 (b)
30 = 15(2) +0
from (b) 15 = 45- 30
from (a) 30 = 210 +45(-4)
hence 15 = 210 (-1) +45 (5)
multiple both sides by 4, we get
60 = 210 (-4) +45 (20)
hence \(x_0 = -4\\x = -4-3t ; t\in [0,14]\)
\(y_0 = 20\\y = 20+14t; t\in[0,14]\)

Answer 3

I don't understand this yet sorry ganeshie can probably help :/

Answer 4

But if I use (a) to solve, I have 30 = 210 +45(-4)
then multiple both sides by 2 to get
60 = 210 (2) +45(-8)
I have different answer with
\(x_0 = -2 \\y_0=-8\)
it confused me a lot.

Answer 5

sorry \(x_0 =2\)

Answer 6

Both are correct

Answer 7

What?? explain me, please

Answer 8

\((2, -8)\) and \((-4, 20)\) are both particular solutions to the diophantine equation \( 210x +45y = 60\)

You just need "any one" particular solution to write out all the solutions.

Answer 9

for example, \((-1, 6)\) is another particular solution

Answer 10

What is the logic?

Answer 11

all the solutions = (any one particular solution) + (null solutions)

Answer 12

null solution is fixed, right? how particular one changes without changing the whole?

Answer 13

Let me ask you a question,
whats the vector equation of the line that passes through \((-4,20)\) and has a direction vector \((-3,14)\) ?

Answer 14

You do scare me, hehehe... I think it is a vector parallel to vector (-3,14) through (-4,20)

Answer 15

write out the equation, im sure you will see the connection once you write out the equaiton

Answer 16

whats the vector equation of the line that passes through \((-4,20)\) and has a direction vector \((-3,14)\) ?

we can express the line as
\[\dbinom{x}{y} = \dbinom{-4}{20} + t\dbinom{-3}{14}\]

yes ?

Answer 17

it is y = -14/3x -4/3

Answer 18

I have asked for equation in "vector" form

Answer 19

ok,

Answer 20

yes, I got this part

Answer 21

look at the vector equation that i wrote
is that the only way to express the equation of line in vector form ?

Answer 22

How about below equation ?
\[\dbinom{x}{y} = \dbinom{2}{-8} + t\dbinom{-3}{14}\]

what line does this represent ? does it represent a different line ?

Answer 23

Yes, they are same.

Answer 24

how do you know they are same ?

Answer 25

just starting point is different, the first one starts at (-4,20), the second one at (2,-8) and both are on the same line.

Answer 26

Exactly! so do you agree that we can use "any one point" on the line to express the line in vector form ?

Answer 27

Yes, I do. :)

Answer 28

good, so do you see how your solution for diophantine equation and the vector form of a line are similar ?

Answer 29

Yes, I got it. Thank you so much. You are.... amazingly patient.

Answer 30

In short :

1) (-4, 20) is a `particular solution` vector

2) all the `null solutions` are given by : t(-3, 14)

3) therefore `all the solutions` are given by (-4, 20) + t(-3, 14)

Answer 31

My midterm is upcoming. I need practice more. Do you know some practice pages online I can use to work with?

Answer 32

which textbook are you using ?

Answer 33

Elementary number theory by Kenneth H. Rosen

Answer 34

I have david m burton
if you want more practice problems on linear diophantine equations, i can take a screenshot of exercise page and send you..

Answer 35

Thank you so much. For this time, I focus on :
1) The fundamental theorem
2)Linear Diophantine
3)Congruences
4) Wilson's theorem , Fermat's theorem and Euler's theorem.

Answer 36

one more: Chinese theorem also

Answer 37

Here are some fun problems.
https://s3.amazonaws.com/upload.screenshot.co/962c284bf3

Solving these on your own gives solid understanding of diophantine equations...

Answer 38

Yes, I will.

Answer 39

Few more
https://s3.amazonaws.com/upload.screenshot.co/c46dd4303d

Answer 40

I will try all. Thanks a lot.