・1K4 $* *$ Find all prime numbers $p$ such that in the base $b$ representation of $p$, each digit $0,1, \cdots$, $b-1$ appears exactly once (0 can lead the number).



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Let $p$ be a prime number that satisfies the condition, and it is represented in base $b$ as $\left(a_{b-1} a_{b-2} \cdots a_{1} a_{0}\right)_{b}$, then $\left\{a_{2}, a_{1}, \cdots, a_{b-1}\right\}=\{0,1, \cdots, b-1\}$,
$$
\begin{aligned}
p & =a_{0}+a_{1} b+a_{2} b^{2}+\cdots+a_{b-1} \cdot b^{b-1} \\
& =\left(a_{0}+a_{1}+\cdots+a_{b-1}\right)+a_{1}(b-1)+a_{2}\left(b^{2}-1\right)+\cdots+a_{b-1}\left(b^{b-1}-1\right) \\
& =\frac{b(b-1)}{2}+a_{1}(b-1)+a_{2}\left(b^{2}-1\right)+\cdots+a_{b-1}\left(b^{b-1}-1\right) .
\end{aligned}
$$

When $b=2 k$, $b-1 \mid p$, so $b=2$. In this case, the only such prime is $2=(10)_{2}$. When $b=2 k+1$, $k \mid p$, so $k=1, b=3$. In this case, it is easy to verify that the only such primes are $(012)_{3}=5,(021)_{3}=7$, $(102)_{3}=11,(201)_{3}=19$.

Therefore, the complete set of primes that satisfy the condition are $2=(10)_{2}, 5=(012)_{3}, 7=(021)_{3}, 11=$ $(102)_{3}, 19=(201)_{3}$.