The 55th question: If a rational number has a finite number of digits in its decimal representation, it is called "short." For positive integers $m, t$, if there exists an integer $c \in\{1,2, \ldots, 2017\}$, such that $\frac{10^{t}-1}{\mathrm{~cm}}$ is short, and for any $\mathrm{k} \in\{1,2, \ldots, \mathrm{t}-1\}, \frac{10^{\mathrm{k}}-1}{\mathrm{~cm}}$ is not short, then $\mathrm{t}$ is called " $\mathrm{m}-$ wonderful." Let $\mathrm{S}(\mathrm{m})$ be the set of $\mathrm{m}-$ wonderful numbers. For $m=1, 2, \ldots$, find the maximum number of elements in $S(m)$.
The 55th question,
Solution: It is easy to know that the rational number $\frac{b}{a}\left(a, b \in Z^{+}, (a, b)=1\right)$ is short if and only if $\mathrm{a}$ does not contain any prime factors other than $2$ and $5$. Therefore, if $m=2^{\alpha} 5^{\beta} s\left(\alpha, \beta \in N, s \in Z^{+}, (s, 10)=1\right)$, then $\frac{10^{t}-1}{m}$ is short if and only if $s \mid (10^{t}-1)$.
Without loss of generality, assume $(m, 10)=1$.
Define $C=\left\{c \in Z^{+} \mid 1 \leq c \leq 2017, (c, 10)=1\right\}$, then $|C|=\left[\frac{2017}{10}\right] \cdot \varphi(10)+3=807$.
According to the definition, the $m-$ wonderful number is the smallest positive integer $t$ such that for some $c \in C$, $10^{t} \equiv 1 \pmod{cm}$. This indicates:
$$
S(m)=\left\{\delta_{\text{cmod}}(10) \mid c \in C\right\}
$$

Here $\delta_{a}(b)$ denotes the order of $b$ modulo $a$. Therefore, for any $m \in Z^{+}$, we have
$|S(m)| \leq |C|=807$.
We will now prove: there exists a positive integer $m$ such that $|S(m)|=807$. That is, we need to prove: there exists a positive integer $m$ such that for all $c \in C, \delta_{cm}(10)$ are distinct.
Let $P=\{p \mid p$ is a prime not exceeding 2017 $\}$.
Let $\alpha=\varphi\left(\prod_{p \mid P} p\right)$, and let $m=10^{\alpha}-1$, then for any $p \in P$, $p \mid m$, and $\delta_{m}(10)=\alpha$.
We first prove a lemma: for each $c \in C$, $\delta_{cm}(10)=c \cdot \alpha$.
Proof of the lemma: Let $\delta_{cm}(10)=\lambda$.
Since $10^{\lambda} \equiv 1 \pmod{cm}$, hence $10^{\lambda} \equiv 1 \pmod{m}$, hence $\alpha \mid \lambda$. Let $\lambda=u \cdot \alpha$. Then $u$ is the smallest positive integer such that $cm \mid (10^{u \cdot \alpha}-1)$.
According to the LTE lemma, we have:
$cm \mid (10^{u \cdot \alpha}-1)$
$\Leftrightarrow v_{p}(cm) \leq v_{p}(10^{u \cdot \alpha}-1)$ (for any $p \in P$)
$\Leftrightarrow v_{p}(c)+v_{p}(m) \leq v_{p}(10^{\alpha}-1)+v_{p}(u)$ (for any $p \in P$)
$\Leftrightarrow v_{p}(c) \leq v_{p}(u)$ (for any $p \in P$)
This shows that the smallest positive integer $u$ satisfying $cm \mid (10^{u \cdot \alpha}-1)$ is $u=c$, hence $\lambda=c \cdot \alpha$. The lemma is proved.
According to the lemma, for such $m$, it satisfies that for all $c \in C, \delta_{cm}(10)$ are distinct. That is, at this time, $|S(m)|=|C|=807$.
In summary, the maximum value of $|S(m)|$ is 807.
Note: If each prime factor of $c$ divides $a^{m}-1$, then $\delta_{c(a^{m}-1)}(a) = a \cdot cm$.