Answer: Option C
$$\eqalign{
& = {17^{200}} \div 18 \cr
& = {\left( {18 - 1} \right)^{200}} \div 18 \cr} $$
Apply Binomial theorem
$$ = {\left( {18} \right)^{200}}{\left( { - 1} \right)^0} + {\left( {18} \right)^{199}}{\left( { - 1} \right)^1} + $$       $$..... + \, {\left( {18} \right)^1}{\left( { - 1} \right)^{199}}$$   $$ \, + \, {\left( {18} \right)^0}{\left( { - 1} \right)^{200}}$$
$$ \Rightarrow $$ Remainder always comes from last term is (18)0 (- 1)200
$$\eqalign{
& = \frac{{{{\left( {18} \right)}^0}{{\left( { - 1} \right)}^{200}}}}{{18}} \cr
& = 1 \times {1^{200}} \cr
& = 1 \cr
& {\text{So, remainder}} = 1 \cr} $$