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.
Best Answer
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 $.00050276520864...$
$$\lfloor\dfrac{10^4}{5}\rfloor = 2000$$
$$\lfloor\dfrac{10^5}{50}\rfloor = 2000$$
$$\lfloor\dfrac{10^6}{502}\rfloor = 1992$$
$$\lfloor\dfrac{10^7}{5027}\rfloor = 1989$$