If  $2^{2006}-2006$  divided by 7, the remainder is  

$2^{2006}-2006=2^2{.2}^{2004}-2006$

$=4\left({\left(2^3\right)}^{668}\right)-2006$

$=4\left({\left(1+7\right)}^{668}\right)-2006$

$=4\left[{668}_{c_0}+{668}_{c_1}.7+{668}_{c_2}{.7}^2+{\mathrm{....668}}_{c_{668}}{.7}^{668}\right]-2006$

$=4+4\left\{{668}_{c_1}\times 7+{668}_{c_2}\times 7^2+\mathrm{....}+{668}_{c_{668}}{.7}^{660}\right\}-2006$

$=-2002+4\times 7\left\{{668}_{c_1}+{668}_{c_2}\times 7+{\mathrm{....7}}^{667}\right\}$

$=7\left\{-286+4\left[{668}_{c_1}+{668}_{c_2}\times 7+\mathrm{....}+7^{667}\right]\right\}$

$\Rightarrow 2^{2006}-2006$ is divisible by 7 with remainder 0