The equation $x^{10}+{(13x-1)}^{10}=0$ has 10 complex roots  $r_1,\overline{r_1},r_2,\overline{r_2},r_3,\overline{r_3},r_4,\overline{r_4},r_5,\overline{r_5},$ where the bar denotes complex conjugation. If the value of  $\frac{1}{r_1\overline{r_1}}+\frac{1}{r_2\overline{r_2}}+\frac{1}{r_3\overline{r_3}}+\frac{1}{r_4\overline{r_4}}+\frac{1}{r_5\overline{r_5}}$ is equal to M then the unit digit of M is 

Divide both sides by  $x^{10}$ to get  $1+{(13-\frac{1}{x})}^{10}=0$

${(13-\frac{1}{x})}^{10}=-1$

Let,  $13-\frac{1}{x}=\omega$ where  $\omega =e^{i(\pi n/5+\pi /10)}$ where $n$ is an integer.
We see that $\frac{1}{x}=13-\omega$. Thus, 

$\frac{1}{x\overline{x}}=(13-\omega )(13-\overline{\omega })=169-13(\omega +\overline{\omega })+\omega \overline{\omega }=170-13(\omega +\overline{\omega })$

Summing over all terms: 

$\frac{1}{r_1\overline{r_1}}+\mathrm{.......}+\frac{1}{r_5\overline{r_5}}=850-13(e^{i\pi /10}+\mathrm{......}+e^{i(9\pi /5+\pi /10)}) =850$