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?
Number of integer solutions of $\frac{1}{x} + \frac{1}{y} = \frac{1}{1000}$
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algebra-precalculus
diophantine-equations
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0@AndréNicolas yes i agree.....49 for positive integer and$98$else. – 2012-02-25
1 Answers
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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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0thnx...got it...49 solutions is the correct answer for positive values of x and y. – 2012-02-25