15. (44th IM() Preliminary Question) Let $m$ be a fixed integer greater than 1, and the sequence $x_{11}, x_{1}, x_{2}, \cdots$ is defined as follows:
$$
x_{i}=\left\{\begin{array}{ll}
2^{i}, & 0 \leqslant i \leqslant m-1, \\
\sum_{j=1}^{m} x_{i j}, & i \geqslant m .
\end{array}\right.
$$

Find the maximum value of $k$ such that there are $k$ consecutive terms in the sequence that are divisible by $m$.
15. Let $r$ be the remainder of $x$ modulo $m$. In the sequence, divide the terms into blocks of $m$ consecutive terms, then the remainders can have at most $m^{nc}$ different cases. By the pigeonhole principle, one type of case will repeat. Since the defined recurrence relation can be extended both forward and backward, the sequence $\{r\}$ is a periodic sequence.
From the given conditions, the backward recurrence formula is
$$
x_{i}=x_{i, \ldots}-\sum_{j=1}^{m i-1} x_{i+}.
$$

The remainders of the $m$ terms are $r_{0}=1, r_{1}=2, \cdots, r_{m-1}=2^{m-1}$. To find the remainders of the $m$ terms preceding these, using the backward recurrence formula, the remainders of the preceding $m$ terms modulo $m$ are $0,0, \cdots, 0,1$. Combining this with 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 $\{r\}$ 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 $m-1$.