The remainder when $2^{2000}$is divided by $17$is

 We have $2^{4 }=16=17–1$ 

$\begin{array}{l}\therefore 2^{2000}={\left(2^4\right)}^{500}=(17-1{)}^{500}\\ {=}^{500}{C}_0{17}^{500}{-}^{500}{C}_1{17}^{499}{+}^{500}{C}_2{17}^{498}-\\ { }^{500}{C}_3{17}^{497}+\dots {-}^{500}{C}_{499}\left(17\right)+(-1{)}^{500}\\ =17m+1\end{array}$

 where$m$ is some positive integer. Thus, the required remainder is $1$.