Mathematics - Basic maths Question with Solution | TestHub

We have $\frac{1}{x}+\frac{1}{y}=\frac{1}{2007}$. This simplifies to $(x-2007)(y-2007)=2007^2=(3^2\cdot 223{)}^2=3^4\cdot 223^2$, which has $(4+1)(2+1)=15$ factors. Since $x<y$, we have $\frac{15-1}{2}=7$ solutions.

FINAL ANSWER: The final answer is $\boxed{7}$