Problem 1022

Three, let $0 . h_{1} h_{2} \cdots$ be a rational number. Then there exist $N_{0}, T \in \mathbf{Z}_{+}$, such that for each $n \geqslant N_{0}$, we have $h_{n+T}=h_{n}$.

First, we prove: there exists $T_{1} \in \mathbf{Z}_{+}, T \mid T_{1}$, and the last non-zero digit of $T_{1}$ is 1.

In fact, let $T=2^{\alpha} \times 5^{\beta} p$, where $\alpha, \beta \in \mathbf{Z}_{+}$, and $p$ is not divisible by 2 and 5. Then
$$T_{0}=2^{\beta} \times 5^{\alpha} T=2^{\alpha+\beta} \times 5^{\alpha+\beta} p=10^{\alpha+\beta} p$$

has the last non-zero digit as an odd number, and not equal to 5.
If it equals 1, then take $T_{1}=T_{0}$;
If it equals 3, then take $T_{1}=7 T_{0}$;
If it equals 7, then take $T_{1}=3 T_{0}$;
If it equals 9, then take $T_{1}=9 T_{0}$.
In the above cases, the last non-zero digit of $T_{1}$ is the same as that of $1, 21, 21, 81$ respectively.
Thus, we find the number $T_{1}$ such that $h_{n+T_{1}}=h_{n}$ when $n \geqslant N$:
$$T_{1}=10^{m}(10 a+1)\left(m, a \in \mathbf{Z}_{+}\right).$$

Next, we prove: for any $n \in \mathbf{Z}_{+}, h_{n} \neq 5$.
In fact, let $[x]$ denote the greatest integer not exceeding the real number $x$. Then in the prime factorization of $n!$, the power index of 2 is
$$r=\left[\frac{n}{2}\right]+\left[\frac{n}{2^{2}}\right]+\cdots,$$
and the power index of 5 is
$$\delta=\left[\frac{n}{5}\right]+\left[\frac{n}{5^{2}}\right]+\cdots.$$

Since for $i \in \mathbf{Z}_{+}$, $\left[\frac{n}{2^{i}}\right] \geqslant\left[\frac{n}{5^{i}}\right]$, we have $r \geqslant \delta$, and
$$n!=2^{\gamma} \times 5^{\delta} q=10^{\delta} \times 2^{\gamma-\delta} q,$$

where $q \in \mathbf{Z}_{+}$, and $q$ is not divisible by 2 and 5. Thus, the last non-zero digit is the same as that of $2^{\gamma-\delta} q$. Therefore, it is not equal to 5. Hence, take a sufficiently large $b \in \mathbf{Z}_{+}$ such that
$$M=10^{m}(10 b+1)>N_{0}.$$

Let $h_{M-1}=h$. Then
$$(M-1)!=10^{k}(10 c+h)\left(c, k \in \mathbf{Z}_{+}\right).$$

Thus, $M!=(M-1)!M$
$$\begin{array}{l} =10^{k}(10 c+h) \cdot 10^{m}(10 b+1) \\ =10^{k+m}[10(b c+h b+c)+h]. \end{array}$$

Therefore, $h_{M}=h$.
Since $T \mid T_{1}$, we have $h_{M-1+T_{1}}=h_{M-1}=h$.
$$\begin{array}{l} \text { Thus, }\left(M-1+T_{1}\right)!=10^{l}(10 d+h)\left(l, d \in \mathbf{Z}_{+}\right). \\ \text { Hence, }\left(M+T_{1}\right)!=\left(M-1+T_{1}\right)!\left(M+T_{1}\right) \\ =10^{l}(10 d+h)\left[10^{m}(10 b+1)+10^{m}(10 a+1)\right] \\ =10^{l+m}(10 d+h)[10(a+b)+2] \\ =10^{l+m}[10(10 a d+10 b d+a h+b h+2 d)+2 h], \end{array}$$

which means $h_{M+T_{1}}$ is the same as the last digit of $2 h$, denoted as $h^{\prime}$.
On the other hand, $h_{M+T_{1}}=h_{M}=h$, but since $h \neq 0.5$, the last digit of $2 h$ is not equal to $h$.
Thus, $h_{M+T_{1}}=h^{\prime} \neq h=h_{M+T_{1}}$, a contradiction.