Note that in $\mod b-1$, the base $10$ representation of the prime number is $p\equiv\frac{(b-1)b}{2}\mod b-1$. If $b$ is even, then we have that $p\equiv 0\mod b-1$. For $p$ to be a prime, we must have $p=b-1$ If $b$ is odd, then $b-1$ is even and $p=k(b-1)+\frac{b-1}{2}=(2k+1)\cdot\frac{b-1}{2}$. This can only potentially be a prime if either $k=0$ or $b=3$. Now we can do casework Case 1: $p=b-1$ Then the base $b$ representation of $p$ is only $1$ digit Case 2: $k=0$ Then we have $p=\frac{b-1}{2}$. Since $p<b$, this will also only have $1$ digit in base $b$. Case 3: $b=3$ Then $p$ must be some permutation of $210_3$. We can test all of them to get that the only solutions are $p=5,7,11,19$