$\text{ If }\left\{\mathrm{x}\right\}\text{ denotes the fractional part of }\mathrm{x}\text{ , then }\left\{\frac{2^{2003}}{17}\right\}\text{ is }$

  $2_{ }{}^{2003}={\left(2^4\right)}^{500}·2^3\Rightarrow 2^{2003}=8(16{)}^{500}\Rightarrow 2^{2003}=8(17-1{)}^{500}$

$\Rightarrow 2^{2003}=8\left[(17{)}^{500}-{}^{500}\mathrm{C}_1(17{)}^{499}+\dots -{}^{500}\mathrm{C}_{499}\left(17\right)+1\right]$

$\frac{2^{2003}}{17}=\text{8·I}+\frac{8}{17},$$where I = {\left(17\right)}^{499} – {}^{500}C_1 {\left(17\right)}^{498} + \dots + {}^{500}C_{499} such that I is an integer.$

$\therefore \left\{\frac{2^{2003}}{17}\right\}=\frac{8}{17}$