Question

The number of positive integer pairs $(x,y)$ such that $\frac{1}{x}+\frac{1}{y}=\frac{1}{2007},x<y$ is

• A. 5
• B. 6
• C. 7
• D. 8

Answer 1

Answer: (C)

7

$\frac{x+y}{xy}=\frac{1}{2007}$

$\Rightarrow xy-2007(x+y)=0$

Adding ${2007}^2$ to both sides, we get

$xy-2007(x+y)+{2007}^2={2007}^2$

$\Rightarrow (x-2007)(y-2007)={2007}^2$

Let $x-2007=A$ and $y-2007=B$

The equation becomes $AB={2007}^2$

Number of solutions of above equation is equal to number of factors of ${2007}^2$

${2007}^2=3^4\times {223}^2$

Hence, number of factors of ${2007}^2$ is $(4+1)(2+1)=15$

In one case $A=B=2007$

Of the remaining $14$ cases, half of the case $A>B$ and remaining half $A<B$

Accordingly we get $7$ cases, where $x<y$.