An equation in two variables? Given 13x + 35y = 2000. How do I find positive integer solutions for this equation (without hit and trial).
My work :- I know I can use Bezout's Theorem to find integer solutions to this equation if I have first solution (x., y.). But I just want positive solutions. To find the first solution I tried using Euclid's algorithm but could not progress further. Please help. Or if you have a smaller way of doing this type of questions feel free to post them as answers. (for all integer solutions or only (+)ve integer solutions). 
 A: $13x+35y=400(70-65)$
$\iff13(x-2000)=35(800-y)$
$\implies \dfrac{13(x-2000)}{35}=800-y $ which is an integer
$\implies35$ divides $13(x-2000)$ hence $x-2000$
$x\equiv2000\pmod{35},x=35t+2000,t>-\dfrac{2000}{35}>-57,$
$13(35t)=35(800-y)\iff y=800-13t>0\iff t<?$
A: It's immediately obvious that $x$ has to be a multiple of $5$, if it has any hope of satisfying this. Hence write $x=5z$ and so we get:
$$35y+65z=2000$$
Now you can apply what Alfred suggested because $35+65=100$ which divides $2000$.
A: The following can only reduce the number of trials and may not exhaust all the solutions.
2000 = (16)(5)(5)(5)
Since a partial factor 5 appears from the ‘35’ and ‘2000’, that means a ’5’ is needed from 13x. Therefore, we should let x = 5h. Then, after cancelling the ‘5’, we have 
13h + 7y = (16)(5)(5)
The RHS indicates there should be at least one more 5 from the LHS. Another thing we can try is let x = 25H instead and at the same time we let y = 5K. Then, 
13H + 7K = 80 which is much easier to guess a possible solution.
The guessing process can be continued if we let H = 4H', ....
A: $\!\bmod 13\!:\ 35y\equiv 2000\!\iff\! -4y\equiv -2$ $\!\iff\! y\equiv \dfrac{1}2\equiv \dfrac{14}2\equiv 7$ $\!\iff x\equiv \dfrac{2000\!-\!35y}{13}\equiv 135$ 
Thus $\,(x,y) = (135,7)+n(-35,13)\,$ has $\,x,y>0\iff 0\le n \le 3$
