Hi

I'm looking for help on solving Linear Diophantine Equation

15y + 27x = 39

partial solution

y = (39-27x)/15

y = (39 - 12x - 15x)/15

y = (39 - 12x)/15 - x

Next step? I'm unsure

Thanks in advance

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- May 15th 2006, 05:02 AMdelpinLinear Diophantine Equations
Hi

I'm looking for help on solving Linear Diophantine Equation

15y + 27x = 39

partial solution

y = (39-27x)/15

y = (39 - 12x - 15x)/15

y = (39 - 12x)/15 - x

Next step? I'm unsure

Thanks in advance - May 15th 2006, 06:04 AMCaptainBlackQuote:

Originally Posted by**delpin**

is the simplest method; but:

First divide through by and rearrange to get:

Now this gives:

So there exists an integer such that:

Now the LHS (left hand side) is even, so must be odd.

So let , then:

But the LHS is divisible by so the RHS must also be divisible by

so must be even., so we may write it as , and then:

Now substituting this back into gives:

So we see that for any :

is a solution.

RonL - May 15th 2006, 11:33 AMThePerfectHacker
Use the following theorem,

Given a diophatine equation with

and is a particular solution

Then all solutions and every solution is,

for an integer

------

You have,

You can leave it the way it is but I suggest to divide by 3,

You need to find a specific solution. Which can be done with trail and error. (Otherwise you can use the Euclidean Algorithm or Coutinued fractions to get a particular solution).

We can easily see that and work.

Also which divides 13.

Thus, all solutions are,

Note it might look different from CaptainBlack's but they are both equivalent. - May 15th 2006, 01:21 PMdelpinThank you
For your help... most appreciated

- May 28th 2006, 08:04 AMSoroban
Hello, delpin!

This is basically Captain Black's and Hacker's solutions ... in baby-talk.

Quote:

I'm looking for help on solving Linear Diophantine Equation:

We have: .

. . Then: .

Multiply both sides by 4:

. .

Hence: . for some integer

Substitute into the original equation: .

. . and we get: .

The solutions are: . for any integer