The equation  $x^{10}+{(11x-1)}^{10}=0$ has 10 complex roots  $r_1,{\overline{r}}_1,r_2,{\overline{r}}_2,r_3,{\overline{r}}_3,r_4,{\overline{r}}_4,r_5,{\overline{r}}_5$. 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 R then  $\frac{R}{122}$ is equal to ____

$$\begin{gathered} x^{10}+{(11x-1)}^{10}=0 \Rightarrow {(11-\frac{1}{x})}^{10}=-1 \\ \Rightarrow consider \omega =11-\frac{1}{x} \\ {\omega }^{10}=-1\Rightarrow \left|\omega \right|=1 \\ \omega =cis\left(\frac{2k\pi +\pi }{10}\right),k=0,1,2------90 \\ \frac{1}{x}=11-\omega \\ \frac{1}{x\overline{x}}=(11-\omega )(11-\overline{\omega })=121+\omega \overline{\omega }-11(\omega +\overline{\omega })=122-11(\omega +\overline{\omega }) \\ \frac{1}{r_1{\overline{r}}_1}+\frac{1}{r_2{\overline{r}}_2}+---+\frac{1}{r_5{\overline{r}}_5}=5.122-11({\omega }_1+{\overline{\omega }}_1+{\omega }_2+{\overline{\omega }}_2+{\omega }_3+{\overline{\omega }}_3+{\omega }_4+{\overline{\omega }}_4+{\omega }_5+{\overline{\omega }}_5) \\ R=5.122 \\ \frac{R}{122}=5 \end{gathered}$$