$\stackrel{\text { Senderov B.A. }}{ }$.

Find the smallest natural number that cannot be represented in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are natural numbers.
$1=\frac{4-2}{4-2}, 2=\frac{8-4}{4-2}, 3=\frac{8-2}{4-2}, 4=\frac{16-8}{4-2}, 5=\frac{32-2}{8-2}$

$6=\frac{16-4}{4-2}, \quad 7=\frac{16-2}{4-2}, 8=\frac{32-16}{4-2}, 9=\frac{128-2}{16-2}, \quad 10=\frac{64-4}{8-2}$.

Assume that $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$. Without loss of generality, let $a>b, c>d$. Denote $m=a-b, n=c-d, k=b-d$. We get

$11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right)$. Since the left side is an odd integer, $k=0$. Note that $n=1$ does not work.

If $m>n>1$, then $2^{m}-1$ and $2^{n}-1$ give a remainder of 3 when divided by 4. Thus, the left and right sides give remainders of 1 and 3, respectively, when divided by 4. This is a contradiction.

## Answer

11.