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Starting with the equation:

$\frac{1}{a}+\frac{1}{b}=\frac{p}{10^n}$,

I reached the equation:

$10^{n-log(p)} = \frac{ab}{a+b}$.

Now given the positive integer $n$, for what integer values of $p$ would the value of:

$10^{n-log(p)}$,

be rational?

Also, given positive integers $n$ and $p$, how would we find positive integer solutions to $a$ and $b$ that satisfy the second equation, where:

$a ≤ b$,

And is it possible to determine, given $n$ and $p$, how many $a$ and $b$ solutions exist?

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    @Khaled Since your $s$econd equation is not really that far removed from the first equation, I'm guessing it would be fairly obvious for the members here who have seen the original PE problem to see the connection. In general, if a user can surreptitiously disguise a PE question and it is answered here and never recognized as a PE question then that user will have only cheated themselves.2012-08-27

2 Answers 2

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As @Qiaochu Yuan noted, $10^{n-\log p}=\dfrac{10^n}{p}$ is always a rational number.



Suppose that a positive integer $n$ and an integr $p$ are given, and let $a$ and $b$ to be integers satisfying the above relation:

$ \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{10^n} \Longleftrightarrow \\ pab = 10^na + 10^nb \Longleftrightarrow \\ p^2ab = 10^npa + 10^npb \Longleftrightarrow \\ p^2ab - 10^npa - 10^npb = 0 \Longleftrightarrow \\ p^2ab - 10^npa - 10^npb + 10^{2n} = 10^{2n} \Longleftrightarrow \\ (pa - 10^n) (pb - 10^n) = 10^{2n} . $


notice that both of the factors $(pa - 10^n)$ and $(pb - 10^n)$ are (positive or negative) divisor of $10^{2n}$, so we can conclude that there is a (positive or negative) divisor $d$ of $10^{2n}$; such that:

$ (pa - 10^n)=d \ \ \ \ \ \ \text{and} \ \ \ \ \ \ (pb - 10^n)=\dfrac{10^{2n}}{d} \ \ \ \ \ \ \Longleftrightarrow $

$ a = \dfrac {d+10^n} {p} \ \ \ \ \ \ \text{and} \ \ \ \ \ \ b = \dfrac { \dfrac{10^{2n}}{d} +10^n} {p} \ \ \ \ \ \ . $

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    @G Cab , Yes; you are right, it must be $10^{2n}$; not $n$. I have been edited that. Thank you so much.2017-08-17
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First of all we shall reduce the RHS fraction, so $ \bbox[lightyellow] { {1 \over a} + {1 \over b} = {p \over {10^{\,n} }} = {{p/\gcd (p,10^{\,n} )} \over {10^{\,n} /\gcd (p,10^{\,n} )}} = {q \over A} }$

Then we have that $ \bbox[lightyellow] { \eqalign{ & {A \over a} + {A \over b} = q\quad \Rightarrow \cr & \Rightarrow \quad \left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor + \left\{ {{A \over a}} \right\} + \left\{ {{A \over b}} \right\} = q\quad \Rightarrow \cr & \Rightarrow \quad \left[ \matrix{ \left( {\left\{ {{A \over a}} \right\} = 0} \right)\; \wedge \;\left( {\left\{ {{A \over b}} \right\} = 0} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q} \right)\quad \quad (1) \hfill \cr \quad \quad \vee \hfill \cr \left( {\left\{ {{A \over a}} \right\} + \left\{ {{A \over b}} \right\} = 1} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q - 1} \right)\quad \quad \quad \;\;(2) \hfill \cr} \right. \cr} }$ where $x = \left\lfloor x \right\rfloor + \left\{ x \right\}$ is the decomposition into floor and fractional part

Now the first condition gives $ \bbox[lightyellow] { \left( {c\backslash A} \right)\; \wedge \;\left( {d\backslash A} \right)\; \wedge \;\left( {\;c + d = q} \right)\quad \Rightarrow \quad \left\{ \matrix{ a = A/c \hfill \cr b = A/d \hfill \cr} \right .\quad\quad (1) }$ i.e., if there are divisors of $A$ summing to $q$, then we can get $a$ and $b$ as thei quotient of $A$ with such divisors.

The second condition translates to $ \bbox[lightyellow] { \left( {{{A\bmod a} \over a} + {{A\bmod b} \over b} = 1} \right)\; \wedge \;\left( {\;\left\lfloor {{A \over a}} \right\rfloor + \left\lfloor {{A \over b}} \right\rfloor = q - 1} \right)\quad \quad \quad \;\;(2) }$ and there is not much more to manage.