Minimum in Floors

Given that $\left \lfloor \dfrac {10^n}x \right \rfloor = 1989$ , where $x, n \in \mathbb N$ . For $x \ge 1$ , implies that $n \ge 5$ . And we have

• For $n=5$ , $x \approx \dfrac {10^5}{1989} \approx 50.2765$ , note that $\begin{cases} \left \lfloor \frac {10^5}{50} \right \rfloor = 2000 \\ \left \lfloor \frac {10^5}{51} \right \rfloor = 1960 \end{cases} \implies \small \color{#D61F06} \text{No solution}$
• For $n=6$ , $x \approx \dfrac {10^6}{1989} \approx 502.765$ , note that $\begin{cases} \left \lfloor \frac {10^6}{502} \right \rfloor = 1992 \\ \left \lfloor \frac {10^6}{503} \right \rfloor = 1988 \end{cases} \implies \small \color{#D61F06} \text{No solution}$
• For $n=7$ , $x \approx \dfrac {10^7}{1989} \approx 5027.65$ , note that $\left \lfloor \dfrac {10^7}{5027} \right \rfloor = \color{#3D99F6}1989 \implies \small \color{#3D99F6} \text{A solution}$

Therefore, the smallest $n$ such that the equation has an integer solution $x$ is $\boxed{7}$ .

Rearranging gives us $\frac{10^n}{k}=x$ where k is between 1989 and 1990. 1/1989 and 1/1990 are .000502765 and .000502513 respectively. We are looking for the first time that an integer is between these values times $10^n$ . We can see from the decimal expansion that this occurs when n=7. (5026 is between 5025.13 and 5027.65)