Given a Diophantine inequality of the form $$a_{1}x_{1}+a_{2}x_{2}+\dots+a_{n}x_{n}\leq N $$ How many positive integer solutions does it have? Here all $a_{i}\geq0$ and $x_{i}\geq 0$.
I was able to find some solutions by considering special cases. Supposing all $x_{i}=0, i\neq 1$. We get, $$a_{1}x_{1}\leq N$$ $$x_{1}\leq\frac{N}{a_{1}}$$ And similarly for other terms we can find that the given Diophantine inequality has at least $$\lfloor\frac{N}{a_{1}}\rfloor+\lfloor\frac{N}{a_{2}}\rfloor+\dots+\lfloor\frac{N}{a_{n}}\rfloor$$ positive integer solutions. I am stuck here and has no idea how to procced forward. I am looking for general method or way to find number of positive solution of any Diophantine inequality. Thank You!