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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
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B
6
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C
7
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D
8
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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 $$x < y$$.


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