(110. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers.
The smallest positive integer sought is 11.
We have:
$$
\begin{array}{c}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2} \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2} \\
2=2 \cdot 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \cdot 5=\frac{2^{6}-2^{2}}{2^{3}-2}
\end{array}
$$

Assume
$$
11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}
$$

Without loss of generality, we can assume $a>b, c>d$. Let $m=a-b, n=c-d, k=b-d$. Then we get
$$
11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right)
$$

The left side of this equation is odd, so $k=0$. It is easy to see that $n=1$ cannot satisfy this equation. If $m>n>1$, then $2^{n}-1$ and $2^{m}-1$ both have a remainder of 3 when divided by 4, which means the left side of the equation has a remainder of 1 when divided by 4, while the right side has a remainder of 3, leading to a contradiction.