Solve This Diophantine Equation : $11x+13y$=$1000$ where $(x,y)$ belongs to positive integers .

Question

Please help me solve this Diophantine Equation : $11x+13y$=$1000$ where $(x,y)$ belongs to positive integers .

Answer 1

I will show you how to find a single solution. General solution can then be found by this. We know that (By Bezout's Lemma) there exist integers $k,l$ such that $$11k+13l=1$$ Now, by trial and error we can easily see that $k = 6$ and $l=-5$. Therefore, $$11(6000)+13(-5000)=1000$$ Now, use a little trick. We want to find positive solution and we are getting negative one for $13$, add and subtract the nearest multiple of $11$ to $5000$ (which is $5005 = 11\times455$), we get, $$11(6000)+13(-5005)+13(5)=1000$$ $$\Longrightarrow11(6000)-13(11)(455)+13(5)=1000$$ $$\Longrightarrow11(6000-13\times455)+13(5)=1000$$ $$\Longrightarrow11(85)+13(5)=1000$$ Hence, we get the required solution.

Hope it is helpful:)

Answer 2

Note that$$y=\frac{1}{13}(1000-11x).$$This is saying that $y$ is an integer when $1000-11x=13k$ for some integer $k$, i.e., when $12\equiv11x\pmod{13}$. This happens when $x\equiv7\pmod{13}$, i.e., when $x=7+13k$ for some integer $k$.

Since $x$ and $y$ have to be positive, $k\geq0$ and$$y=\frac{1}{13}(1000-11(7+13k))>0\implies k<\frac{71}{11}\implies k\leq6.$$So there are $7$ solutions and they are $(7,71)$, $(20,60)$, $(33,49)$, $(46,38)$, $(59,27)$, $(72,16)$, and $(85,5)$.