Question

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

A
5
B
6
C
7
D
8

Solution

The correct option is B 7$$\dfrac{x+y}{xy}=\dfrac{1}{2007}$$$$\Rightarrow xy-2007(x+y)=0$$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 $$x < y$$.Maths

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