Solutions to ceiling equation system 
Prove that there does not exist an $x$ with $1000 \leq x \leq 1990$ that can be expressed in the forms $$\dfrac{10000}{x} = \left \lceil\dfrac{10000}{x} \right \rceil -\dfrac{1}{m} \quad \text{and} \quad \dfrac{10000}{x-1} = \left \lceil\dfrac{10000}{x-1} \right \rceil-\dfrac{1}{k}$$ where $m$ and $k$ are positive integers and $\left \lceil\dfrac{10000}{x} \right \rceil = \left \lceil\dfrac{10000}{x-1} \right \rceil$.

I wasn't sure how to go about solving this question. I found an example that worked but instead of $10000$ it was $1000$. We have $$\dfrac{1000}{144} = 7-\dfrac{1}{18} \quad \text{and} \quad \dfrac{1000}{143} = 7-\dfrac{1}{143}.$$ How do we go about finding solutions here?
 A: Some necessary conditons can be found, for example both $x \cdot 10000 \bmod x$ and  $(x-1) \cdot 10000 \bmod (x-1)$ must be even multiples of triangular numbers. Unfortunately I did not see how to use those for a systematic approach to find solutions. (Not that it matters, but I am curious what the source or background of the problem is.)
By brute force, there are no solutions $1000 \le x \le 1990$. Dropping the range constraint, however, there exists a set of solutions that work for arbitrary $N = 10^n$.


*

*$n = 2$
$$\frac{100}{34} = 3 - \frac{1}{17}, \;\;\; \frac{100}{35} = 3 - \frac{1}{7}$$

*$n = 3$
$$\frac{1000}{334} = 3 - \frac{1}{167}, \;\;\; \frac{1000}{335} = 3 - \frac{1}{67}, \;\;\; \frac{1000}{336} = 3 - \frac{1}{42}$$

*$n = 4$
$$\frac{10000}{3334} = 3 - \frac{1}{1667}, \;\;\; \frac{10000}{3335} = 3 - \frac{1}{667}, \;\;\; \frac{10000}{3336} = 3 - \frac{1}{417}$$

*$...$
The explicit solution is $x = \frac{10^n + 5}{3}$ for $n \ge 2$, and $(x+1)$ is a solution as well for $n \ge 3$.
The proof follows by simply substituting the values and noting that $5 \,|\, x$, $2 \,|\,x-1$, and (for $n \ge 3$) also $8 \,|\, x+1$.
$$\frac{10^n}{x-1} = \frac{3 \cdot 10^n}{10^n + 2} = 3 - \frac{6}{10^n + 2} = 3 - \frac{1}{(\frac{x-1}{2})}
$$
$$\frac{10^n}{x} = \frac{3 \cdot 10^n}{10^n + 5} = 3 - \frac{15}{10^n + 5} = 3 - \frac{1}{(\frac{x}{5})}
$$
$$\frac{10^n}{x+1} = \frac{3 \cdot 10^n}{10^n + 8} = 3 - \frac{24}{10^n + 8} = 3 - \frac{1}{(\frac{x+1}{8})}
$$
A: Denote $\lceil\frac{10000}{x}\rceil=A$ and $\lceil \frac{10000}{x-1}\rceil=B$. We have $1000\le x\le 1990$.
A tedious calculation gives five values to be discarded because $A\ne B$ and five sets of candidates to be considered:  $$\begin{cases}x=1000 \Rightarrow A=10\text{ and } B=11\\x=1012\Rightarrow A=9\text{ and } B=10\\x=1250\Rightarrow A=8\text{ and } B=9\\x=1429\Rightarrow A=7\text{ and } B=8\\x=1667\Rightarrow A=6\text{ and } B=7\end{cases}$$ 
$$\begin{cases}1001\le x\le 1111\Rightarrow A=B=10\\1013\le x\le 1249\Rightarrow A=B=9\\1251\le x\le 1428\Rightarrow A=B=8\\1430\le x\le 1666\Rightarrow A=B=7\\1668\le x\le 1990\Rightarrow A=B=6\end{cases}$$
Denoting $a=\lceil\frac{10000}{x}\rceil$ we need $$\frac Nx=a-\frac1m\\\frac{N}{x-1}=a-\frac 1k$$ which implies $$1=\frac{N(k-m)}{(am-1)(ak-1)}\qquad (*)$$
Among the five possible values  for $a$, there is not solution for  $a=10,8,6$ (almost visible because the corresponding denominators in $(*)$ become odd) Hence we get two necessary conditions: $$1=\frac{10000(k-m)}{(9m-1)(ak-1)}\qquad (**)\\1=\frac{10000(k-m)}{(7m-1)(7k-1)}\qquad (***)$$ 
The equations $(**)$ and $(***)$  have the solutions $(m,k)=(139,-1111),(-1111,-111)$ and $(x,y)=(268,-857),(3393,-217),(199,7768)$ respectively.
These necessary solutions of $(*)$, calculation shows are not compatible for the problem.
Thus there is not solution.
