5. Suppose 1993 can be written as a three-digit number $\overline{x y z}$ in base $b$, and $x+y+z=1993$. Determine all possible values of $x, y, z$ and $b$.


5. From the given, we have $\left\{\begin{array}{l}x \cdot b^{2}+y \cdot b+z=1993, \\ x+y+z=22 .\end{array}\right.$
(1) - (2) gives $(b-1)[(b+1) x+y]=1971$, so $(b-1)$ is a divisor of $1971=3^{3} \cdot 73$.
Also, $b^{3}>1993, b^{2}<1993$, so $12<b<45$, thus $b-1=3^{3}=27$, which means $b=28$.
Furthermore, $1993=2 \cdot 28^{2}+15 \cdot 28+5$, hence $x=2, y=15, z=5, b=28$.