Team selection tests for BMO 2018

Let $S$ be the set of positive integers which can be written as $s=\frac{2^a-2^b}{2^c-2^d}$ for some positive integers $a,b,c,d$. Since $s>0$, we can assume that $a>b,c>d$ and write $s=2^{b-d}\frac{2^{a-b}-1}{2^{c-d}-1}$. It's now clear that $b-d=v_2(x)$. So if we set $x=2^{v_2(x)}y$ then $y=\frac{2^u-1}{2^v-1}$ ($u,v$ are positive integers). In particular, the number we are looking for is odd. Clearly the numbers of the form $2^u-1$ ($v=0$) belong to $S$. More generally, we know that $2^v-1\mid 2^u-1$ if and only if $v\mid u$. Set $u=dv$ then $$y=\frac{2^u-1}{2^v-1}=\frac{2^{dv}-1}{2^v-1}=2^{(d-1)v}+2^{(d-2)v}+\cdots +2^v+1$$ This means that the base 2 representation of $y$ is of the form $$y=1\underset{v-1}{\underbrace{0\dots 0}}1\underset{v-1}{\underbrace{0\dots 0}}1\dots \underset{v-1}{\underbrace{0\dots 0}}1$$ ($d$ digits 1, every two consecutive 1's are separated by $v-1$ 0's). So the number we are looking for is the smallest odd number not of this form (for some $d\ge 1,v\ge 1$). Note that $$1=\overline{1},3=\overline{11},5=\overline{101},7=\overline{111},9=\overline{1001},11=\overline{1011}$$ We can conclude that 11 is the smallest odd number whose base 2 representation is not of the desired form. So the answer is $n=11$.