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A rational number is called "lucky" if it equals both $a+\frac{b}{c}$ and $a\times\frac{b}{c}$ for some positive integers $a,b,c$. How many lucky numbers are there between $5$ and $10$?

Here's what I have so far: $$a+\frac{b}{c}=a\times\frac{b}{c}$$ $$\frac{ac}{c}+\frac{b}{c}=\frac{ab}{c}$$ $$\frac{ac+b}{c}=\frac{ab}{c}$$ $$ac+b=ab$$ I don't know what to do after this. Any ideas?

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    $\begingroup$ Immediate observation : $$a(b-c)=b\iff \frac ba=b-c$$ $\endgroup$ – lab bhattacharjee May 18 '14 at 5:51
  • $\begingroup$ And note b has to be greater than c meaning there aren't many values that a can take $\endgroup$ – Jack Yoon May 18 '14 at 5:56
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Solving $a+r=ar$ (where $r=b/c$) gives $r=a/(a-1)$, and the corresponding lucky number is $ar=a^2/(a-1)$. For $a=2,3,4,5,6,7,8,9,\ldots$, this is $4,9/2,16/3,25/4,36/5,49/6,64/7,81/8,\ldots$. The sequence then continues to increase. So there are just five lucky numbers between $5$ and $10$.

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