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!
Best Answer
The trick is to realize $7\times0.7143=5.0001$.
First let's multiply by $10^n$ to shift the decimal to the right - hence for some integers $k$ and $n$ and real number $c\in[0,1)$ we can write
$$\frac{10^na}{b}=k+0.7143+0.0001c$$
Now multiply by our magic number!
$$7\times\frac{10^na}{b}=7k+5.0001+0.0007c$$
Hence
$$7(10^na-kb)-5b=(0.0001+0.0007c)b$$
The left hand side is an integer, so there exists $m\in\mathbb{Z}$ such that
$$m=(0.0001+0.0007c)b$$
Since $0\le c<1$, we have
$$0.0001 b\le m<0.0008b$$
Additionally, since $b\in\mathbb{N}$, we have $$0<0.0001b\le m$$ so because $m$ is an integer, $$1\le m<0.0008b$$ giving $1250<b$ as desired.