20. N1 (POL) Let \( m \) be a fixed integer greater than 1. The sequence \( x_{0}, x_{1}, x_{2}, \ldots \) is defined as follows:  
\[ x_{i}= \begin{cases}2^{i}, & \text { if } 0 \leq i \leq m-1 \\ \sum_{j=1}^{m} x_{i-j}, & \text { if } i \geq m\end{cases} \]  
Find the greatest \( k \) for which the sequence contains \( k \) consecutive terms divisible by \( m \).
20. Let $r_{i}$ be the remainder when $x_{i}$ is divided by $m$. Since there are at most $m^{m}$ types of $m$-consecutive blocks in the sequence $\left(r_{i}\right)$, some type will repeat at least twice. Then since the entire sequence is determined by one $m$-consecutive block, the entire sequence will be periodic. The formula works both forward and backward; hence using the rule $x_{i}=$ $x_{i+m}-\sum_{j=1}^{m-1} x_{i+j}$ we can define $x_{-1}, x_{-2}, \ldots$. Thus we obtain that 
$$ \left(r_{-m}, \ldots, r_{-1}\right)=(0,0, \ldots, 0,1) $$
Hence there are $m-1$ consecutive terms in the sequence $\left(x_{i}\right)$ that are divisible by $m$. If there were $m$ consecutive terms in the sequence $\left(x_{i}\right)$ divisible by $m$, then by the recurrence relation all the terms of $\left(x_{i}\right)$ would be divisible by $m$, which is impossible.