Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$

Question

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.

Answer 1

$\frac{10^4}5=2000>1989$

$\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:

$\lfloor{\frac {10^7}{5027}}\rfloor=1989$

Answer 2

Rewrite the equation as

$$\frac{10^n}{x} = 1989+\epsilon$$

where $0 < \epsilon < 1$ and $x$ is a positive integer.

Then $\dfrac{10^n}{1989 + \epsilon} = x$ and

$\dfrac{10^n}{1989}$ must be slightly larger than an integer.

The first few digits of $\dfrac{1}{1989}$ are $.00050276520864...$

$$\lfloor\dfrac{10^4}{5}\rfloor = 2000$$

$$\lfloor\dfrac{10^5}{50}\rfloor = 2000$$

$$\lfloor\dfrac{10^6}{502}\rfloor = 1992$$

$$\lfloor\dfrac{10^7}{5027}\rfloor = 1989$$