<s>[INST] Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$ 
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$.

My work-

$\frac{10^n}{x}-1<\lfloor \frac{10^n}{x}\rfloor≤\frac{10^n}{x}\Rightarrow\frac{10^n}{x}-1<1989≤\frac{10^n}{x}\Rightarrow\frac{10^n}{1990}<x≤\frac{10^n}{1989}$

I am unable to proceed beyond this. Any help or other method is appreciated.
 [/INST] $\frac{10^4}5=2000>1989$
$\Rightarrow x>5 $ for n=4
let $x=5.1$ for $10^4$ or 51 for $10^5$ we have:
$\frac{10^5}{51}=1960$
$\Rightarrow 5.1>x>5$ for $n=5$
let $x=50.1$ for $10^5$ or 501 for $10^6$ we have:
$\frac{10^6}{501}=1996$
$\Rightarrow  502>x>501$ for n=6
let $x=502$ we have:
$\frac {10^6}{502}=1992$
$\Rightarrow 503>x>502 $ for $n=6$ or $5020<x<5030$ for $10^7$
With few try we find $x=5027$ is the integer solution for $n=7$:
$\frac{10^7}{5027}=1989.25$
or:
$\lfloor{\frac {10^7}{5027}}\rfloor=1989$
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