17. (6 points) The remainder of $2^{2017} \div 7$ is $\qquad$ . (Note: $x^{n}$ represents $n$ $x$s multiplied together)
【Solution】Solution: According to the analysis, transform $2^{2017} \div 7$ into another form, that is: $2^{2017} \div 7=2 \times 2^{2016} \div 7=2 \times 2^{16}$
$$
\begin{array}{l}
\times 126 \div 7=2 \times\left(2^{16}\right)^{126} \div 7, \\
\because 2^{16}=4096=7 \times 585+1,
\end{array}
$$
$\therefore 2^{16} \div 7$ has a remainder of 1, by the remainder theorem, $\left(2^{16}\right)^{126} \div 7$ has a remainder of $1^{126}=1$, then $2 \times\left(2^{16}\right)^{126} \div 7$ has a remainder of: $2 \times 1=2$, that is, $2^{2017} \div 7$ has a remainder of 2.

Therefore, the answer is: 2.