Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \in\{1,2,3, \ldots, 2017\}$ such that $\frac{10^{t}-1}{c \cdot m}$ is short, and such that $\frac{10^{k}-1}{c \cdot m}$ is not short for any $1 \leqslant k<t$. Let $S(m)$ be the set of $m$-tastic numbers. Consider $S(m)$ for $m=1,2, \ldots$. What is the maximum number of elements in $S(m)$ ? (Turkey) Answer: 807.
First notice that $x \in \mathbb{Q}$ is short if and only if there are exponents $a, b \geqslant 0$ such that $2^{a} \cdot 5^{b} \cdot x \in \mathbb{Z}$. In fact, if $x$ is short, then $x=\frac{n}{10^{k}}$ for some $k$ and we can take $a=b=k$; on the other hand, if $2^{a} \cdot 5^{b} \cdot x=q \in \mathbb{Z}$ then $x=\frac{2^{b} \cdot 5^{a} q}{10^{a+b}}$, so $x$ is short. If $m=2^{a} \cdot 5^{b} \cdot s$, with $\operatorname{gcd}(s, 10)=1$, then $\frac{10^{t}-1}{m}$ is short if and only if $s$ divides $10^{t}-1$. So we may (and will) suppose without loss of generality that $\operatorname{gcd}(m, 10)=1$. Define 
$$ C=\{1 \leqslant c \leqslant 2017: \operatorname{gcd}(c, 10)=1\} . $$ 
The $m$-tastic numbers are then precisely the smallest exponents $t>0$ such that $10^{t} \equiv 1$ $(\bmod \mathrm{cm})$ for some integer $c \in C$, that is, the set of orders of 10 modulo cm. In other words, 
$$ S(m)=\left\{\operatorname{ord}_{c m}(10): c \in C\right\} $$ 
Since there are $4 \cdot 201+3=807$ numbers $c$ with $1 \leqslant c \leqslant 2017$ and $\operatorname{gcd}(c, 10)=1$, namely those such that $c \equiv 1,3,7,9(\bmod 10)$, 
$$ |S(m)| \leqslant|C|=807 $$ 
Now we find $m$ such that $|S(m)|=807$. Let 
$$ P=\{10\}. We will show that $k=c$. Denote by $\nu_{p}(n)$ the number of prime factors $p$ in $n$, that is, the maximum exponent $\beta$ for which $p^{\beta} \mid n$. For every $\ell \geqslant 1$ and $p \in P$, the Lifting the Exponent Lemma provides 
$$ \nu_{p}\left(10^{\ell \alpha}-1\right)=\nu_{p}\left(\left(10^{\alpha}\right)^{\ell}-1\right)=\nu_{p}\left(10^{\alpha}-1\right)+\nu_{p}(\ell)=\nu_{p}(m)+\nu_{p}(\ell) $$ 
so 
$$ \begin{aligned} c m \mid 10^{k \alpha}-1 & \Longleftrightarrow \forall p \in P ; \nu_{p}(c m) \leqslant \nu_{p}\left(10^{k \alpha}-1\right) \\ & \Longleftrightarrow \forall p \in P ; \nu_{p}(m)+\nu_{p}(c) \leqslant \nu_{p}(m)+\nu_{p}(k) \\ & \Longleftrightarrow \forall p \in P ; \nu_{p}(c) \leqslant \nu_{p}(k) \\ & \Longleftrightarrow c \mid k . \end{aligned} $$ 
The first such $k$ is $k=c$, so $\operatorname{ord}_{c m}(10)=c \alpha$. 

Comment. The Lifting the Exponent Lemma states that, for any odd prime $p$, any integers $a, b$ coprime with $p$ such that $p \mid a-b$, and any positive integer exponent $n$, 
$$ \nu_{p}\left(a^{n}-b^{n}\right)=\nu_{p}(a-b)+\nu_{p}(n), $$ 
and, for $p=2$, 
$$ \nu_{2}\left(a^{n}-b^{n}\right)=\nu_{2}\left(a^{2}-b^{2}\right)+\nu_{p}(n)-1 . $$ 
Both claims can be proved by induction on $n$.