Sequence of digits $7143$ appears somewhere after the comma Let $m,n$ be positive integers such that in the decimal representation of $\frac{m}{n}$, the sequence of digits $7143$ appears somewhere after the comma. Prove that $n>1250$.
For small values of $n$ we can check all possible sequence of digits appearing after the comma by writing out $\frac{1}{n},\frac{2}{n},\cdots,\frac{n-1}{n}$, but doing that up to $n=1250$ without a computer would be too much. 
 A: The machinery of continued fractions will give this, though not very elegantly. Here's a sketch: Observe first of all that we may assume that
$$
\frac{m}{n}=0.7143...= \frac{7143+q}{10^4} ,
$$
with $0\le q<1$. Now run the continued fraction algorithm, as described in the link, on this number: we obtain that $a_0=0$,
$$
\frac{10^4}{7143+q}= 1+ \frac{2857-q}{7143+q} ,
$$
(so $a_1=1$), then
$$
\frac{7143+q}{2857-q} = 2 + \frac{1429+3q}{2857-q} ,
$$
so $a_2=2$, then
$$
\frac{2857-q}{1429+3q} = 1 + \frac{1428-4q}{1429+3q} ,
$$
so $a_3=1$, then $a_4=1$, and the punchline is that $a_5$ will be huge. If we work it out carefully, we obtain that $a_5\ge 178$ (the worst case arises when $q\simeq 1$).
Since our number $m/n$ is rational, the algorithm terminates at some point, let's say at step $N$, and then $m/n$ equals the convergent $h_N/k_N$. The denominators satisfy the recurrence relation $k_n=a_n k_{n-1}+k_{n-2}$, with initial values $k_{-2}=1$, $k_{-1}=0$ (see again the wikipedia article). This successively gives $k_0=1$, $k_1=1$, $k_2=3$, $k_3=4$, $k_4=7$, $k_5\ge 178\cdot 7+ 4 = 1250$.
We'd need $q=1$ for equality, and then indeed $7144/10^4=893/1250$, but $q=1$ was not allowed above because this number doesn't have the required $3$ in the $4$th digit (it equals $0.7144$). So we also obtain strict inequality.
I suspect there's a simpler solution, using the calculation from the last paragraph.
