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What is the number of integer solutions of: $\frac{1}{x} + \frac{1}{y} = \frac{1}{1000}$ How to solve these type of problems if am comfortable of solving $x+y=z$. But how to do if multiplicative inverses are involved?

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    @AndréNicolas yes i agree.....49 for positive integer and$98$else.2012-02-25

1 Answers 1

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Assuming you mean integer solutions, you will be able to rewrite your equation as:

$1000(x+y) = xy$

Then rearranging you will be able to write as:

$(x - 1000)(y - 1000) = 1000^2$

So that your solutions for $x-1000$ and $y-1000$ correspond to divisors of $1000^2$.

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    thnx...got it...49 solutions is the correct answer for positive values of x and y.2012-02-25