smc

We are going to look at this problem in binary. $x_0=(0.a_1{a}_2\cdots {)}_2$ $2x_0=(a_1.a_2{a}_3\cdots {)}_2$ If $2x_0<1$, then $x_0<\frac{1}{2}$ which means that $a_1=0$ and so $x_1=(.a_2{a}_3{a}_4\cdots {)}_2$ If $2x_0\ge 1$ then $x\ge \frac{1}{2}$ which means that $x_1=2x_0-1=(.a_2{a}_3{a}_4\cdots {)}_2$. Using the same logic, we notice that this sequence cycles and that since $x_0=x_5$ we notice that $a_n=a_{n+5}$. We have $2$ possibilities for each of $a_1$ to $a_5$ but we can't have $a_1=a_2=a_3=a_4=a_5=1$ so we have $2^5-1=\boxed{\text{31}}$