18.4.3 $\star \star$ A positive integer $n$ is not divisible by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
Parse $1=\left|2^{1}-3^{1}\right|, 5=\left|2^{3}-3^{1}\right|, 7=\left|2^{1}-3^{2}\right|, 11-\left|2^{4}-3^{3}\right|, 13=$ $\left|2^{4}-3^{1}\right|, 17=\left|2^{6}-3^{4}\right|, 19=\left|2^{3}-3^{3}\right|, 23=\left|2^{5}-3^{2}\right|, 25=\left|2^{1}-3^{3}\right|$, $29=\left|2^{5}-3^{1}\right|, 31=\left|2^{5}-3^{0}\right|$.
We prove, for non-negative integers $a, b,\left|2^{a}-3^{b}\right|=35$ does not hold.
(1) $2^{a}-3^{b}=35$. Clearly $a \geqslant 6 . \bmod 8$ gives, $3^{b}=-3(\bmod 8) \cdots$ (1), but $3^{b}=1$ or 3 $(\bmod 8)$. Therefore (1) does not hold.
(2) $3^{b}-2^{a}=35 \cdots$ (2). Clearly $b \geqslant 1,2^{a} \geqslant 81-35-46, a \geqslant 6 . \bmod 4$ gives, $(-1)^{b} \equiv-1(\bmod 4)$, so $b$ is an odd number $2 k+1$.
(5) $\bmod 7$ gives, $3 \times 2^{k} \equiv 2^{a}(\bmod 7)$. That is, $2^{a-k}=3(\bmod 7) \cdots$ (3). But $2^{2} \equiv 4,2^{3} \equiv 1$, $2^{3 k+1} \equiv 2,2^{3 k+2} \equiv 4,2^{3 k} \equiv 1,(\bmod 7)$. Therefore (3) does not hold.
Thus, the smallest value of $n$ is 35.