7. Find the smallest positive integer which cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ where $a, b, c, d$ are non-negative integers.
(1 mark)
求最小的正整數, 該數不能表達成 $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ （其中 $a 、 b 、 c 、 d$ 為非負整數）的形式。
(1 分)
7. 11
7. Call a positive integer 'good' if it can be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ where $a, b, c, d$ are non-negative integers. Clearly, if $n$ is 'good', then so is $2 n$ because we can simply increase $a$ and $b$ by 1 to double the value of $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$.
Note that $1,3,5,7,9$ are 'good' since $1=\frac{2^{2}-2^{1}}{2^{2}-2^{1}}, 3=\frac{2^{3}-2^{1}}{2^{2}-2^{1}}, 5=\frac{2^{4}-2^{1}}{2^{2}-2^{0}}, 7=\frac{2^{4}-2^{1}}{2^{2}-2^{1}}$ and $9=\frac{2^{6}-2^{0}}{2^{3}-2^{0}}$. Hence $2,4,6,8,10$ are also 'good' by the remark in the previous paragraph.
Finally, assume $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}=\frac{2^{k}\left(2^{m}-1\right)}{2^{n}-1}$ where $m=a-b, n=c-d$ and $k=b-d$, with $m, n$ positive. It follows that $11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right)$. Since the left hand side is odd, we have $k=0$. Clearly, neither $m$ nor $n$ can be equal to 1 . Thus $2^{m}-1 \equiv 2^{n}-1 \equiv 3(\bmod 4)$. As $11 \equiv 3(\bmod 4)$, we get a contradiction as the left hand side is congruent to 1 but the right hand side is congruent to 3 modulo 4 . Thus 11 is not 'good' and so the answer is 11 .