1. Let $m$ be a fixed integer greater than 1, and the sequence $x_{0}, x_{1}$, $x_{2}, \cdots$ is defined as follows:
$$
x_{i}=\left\{\begin{array}{ll}
2^{i}, & 0 \leqslant 1 \leqslant m-1, \\
\sum_{j=1}^{m} x_{i-j}, & i \geqslant m .
\end{array}\right.
$$

Find the maximum value of $k$ such that there are $k$ consecutive terms in the sequence that are divisible by $m$.
(Poland provided)
Solution: Let $r_{i}$ be the remainder of $x_{i}$ modulo $m$. In the sequence, divide the terms into blocks of $m$ consecutive terms. Then, there are at most $m^{m}$ different cases for the remainders. By the pigeonhole principle, one type of case will appear repeatedly. Since the defined recurrence relation can be used to deduce both forward and backward, the sequence $\left\{r_{i}\right\}$ is a periodic sequence.
From the given conditions, the backward recurrence formula is
$$
x_{i}=x_{i+m}-\sum_{j=1}^{m-1} x_{i+j} \text {. }
$$

The remainders of the $m$ terms are $r_{0}=1, r_{1}=2$, $\cdots, r_{m-1}=2^{m-1}$. To find the remainders modulo $m$ of the $m$ terms preceding these, using the backward recurrence formula, the remainders of the preceding $m$ terms modulo $m$ are $\underbrace{0,0, \cdots, 0,1}_{m-1}$. Combining the periodicity of the remainder sequence, we get $k \geqslant m-1$.

On the other hand, if there are $m$ consecutive terms in the remainder sequence $\left\{r_{i}\right\}$ that are all 0, then by the backward and forward recurrence formulas, for all $i \geqslant 0$, we have $r_{i}=0$, which is a contradiction.
Therefore, the maximum value of $k$ is $m-1$.