TITLE: If the decimal expansion of $a/b$ contains "$7143$" then $b>1250$
QUESTION [41 upvotes]: I recently stumbled upon this really interesting problem:

Suppose we have a fraction $\frac{a}{b}$ where $a,b \in \mathbb{N}$ and we know that the decimal fraction of $\frac{a}{b}$ has the numerical sequence $7143$ somewhere in the decimal place. Show that $b > 1250 $.

This question is part of der Bundeswettbewerb Mathematik 2015, zweite Rund. The competition ended September 1st, 2015.
Any kind of help will be appreciated!

REPLY [6 votes]: By multiplying $a$ with a power of $10$ if necessary, we may assume that $1743$ occurs immediately after the decimal point, and by subtacting a multiple of $b$ from $a$ we may assume that in fact $\frac ab=0.1743\ldots$.
Then 
$$ \frac57=0.7142857\ldots <\frac ab <0.7145 = \frac{893}{1250}$$
Hence the numeraotrs of the difference fractions $\frac ab-\frac 57=\frac{7a-5b}{7b}$ and $\frac{893}{1250}-\frac ab=\frac{893b-1250a}{1250b}$ are positive integers, i.e.,
$$ \begin{align}7a-\hphantom{88}5b&\ge 1\\-1250a+893b&\ge1\end{align}$$
Then $1250$ times the first plus $7$ times the second inequality eliminates $a$ and gives us
$$ (-5\cdot 1250+7\cdot 893) b\ge 1250+7,$$
i.e.,
$$ b\ge 1257.$$