Déjà vu?

WLOG we can assume $a > b$ and $c > d$ (we can switch them if necessary, and the quotient has to be positive). Now note that $b \ge d$ because otherwise the quotient is not an integer. Divide through by $2^d$ to get something of the form $2^x \frac{2^y-1}{2^z-1}$ , where $x$ is nonnegative and $y,z$ are positive.

It's clear that this is an integer if and only if $z|y$ . In that case, we can rewrite as $2^x(1+2^z+2^{2z}+\cdots+2^{y-z})$ . So the integers that can be written as a quotient like this are precisely the integers which, when we write their binary expansion, have their $1$ 's spaced evenly apart.

To get the smallest integer that fails this criterion, we clearly need at least three $1$ 's in the expansion, and since $111$ works (with $z = 1$ and $y = 3$ ), we need at least four binary digits. This leads to $1101$ and $1011$ ; the smaller of these, $1011$ , is the decimal number $\fbox{11}$ .

Call a positive integer $n$ good if it can be expressed as $\dfrac{2^a-2^b}{2^c-2^d}$ for some positive integers $a,b,c,d.$

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We first make the following observations.

• All powers of $2$ are good.

Proof: Note that $2^n = \dfrac{2^{n+2} - 2^{n+1}}{2^2 - 2}.$ $\blacksquare$

• All numbers of the form $2^k-1$ are good.

Proof: Note that $2^k - 1 =\dfrac{2^{k+1} - 2}{2^2-2}.$ $\blacksquare$

• All numbers of the form $2^k+1$ are good.

Proof: Note that $2^k+1 = \dfrac{2^{2k+1} - 2}{2^{k+1} - 2}.$ $\blacksquare$

• $n$ is good if and only if its largest odd factor is good.

Proof: Let $j$ be the largest odd factor of $n,$ and let $n= 2^m j.$ Let $n= \dfrac{2^a- 2^b}{2^c-2^d}.$ Then, $j= \dfrac{2^a-2^b}{2^{c+m} - 2^{d+m}}.$ Conversely, if $j = \dfrac{2^p - 2^q}{2^r-2^s}$ is good, so is $n = 2^mj = \dfrac{2^{p+m} - 2^{q+m}}{2^r - 2^s}.$ $\blacksquare$

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By the above observations, we immediately reject $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$ We shall now show that $11$ is not a good number. Suppose $11 = \dfrac{2^a-2^b}{2^c-2^d},$ where $a>b$ and $c>d.$ Rearrange it as $2^b (2^{a-b} - 1) = 11 \times 2^d (2^{c-d} - 1).$ Matching the highest powers of $2$ in both sides, $b=d.$ Then, the equation becomes $2^b (2^{a-b}-1) =11 \times 2^b (2^{c-b} - 1).$ Again, rewrite the equation as $2^{a-b} + 10 = 11 \times 2^{c-b}.$ Note that $c-b \leq 1,$ since otherwise we would have that $4 \mid 11 \times 2^{c-b}$ and $4 \mid 2^{a-b} ,$ which implies $4 \mid 10.$ Again, since $c>d=b,$ $c-b \neq 0.$ So it follows that $c-b=1.$ Our equation becomes $2^{a-b} = 12,$ not possible.

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In conclusion, the answer is $\boxed{11}.$

2, 4, 6, 8, 12, 14, 16, 24, 28, 30,32, 48, 50, 56, 60, 62, 64, 92, 96, ...; multiplied with {11, 13, 19, 22, 26, ...} cannot be on them; 11 is hence the smallest answer.

C.K. Lu, 40, Malaysia