# 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}

I am unable to proceed beyond this. Any help or other method is appreciated.

• Perhaps a good idea would be to try this with a number smaller than $1989$.
– lulu
Commented Oct 13, 2021 at 15:08
• Should say, a simple search only takes a few seconds (with a calculator, of course).
– lulu
Commented Oct 13, 2021 at 15:14
• $10^7/5027=1989.25...$. I do not think n can be less than 7. Commented Oct 13, 2021 at 15:55

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 $$.00050276520864...$$

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

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

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

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

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

$$\Rightarrow x>5$$ 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 $$5020 for $$10^7$$

With few try we find $$x=5027$$ is the integer solution for $$n=7$$:

$$\frac{10^7}{5027}=1989.25$$

or:

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