1. Let the decimal number 1999 be written as a three-digit number $\overline{x y z}$ in base $b$, and $x+y+z=1+9+9+9$. Then the value of $b$ is ( ).
(A)11
(B) 15
(C) 18
(D) 28
$-1 . \mathrm{D}$.
[Solution 1]: From the problem, we have
$$
x b^{2}+y b+z=1999 .(x \geqslant 1)
$$

Thus, $b^{3}>1999, b^{2} \leqslant 1999$.
Therefore, $12<b<45$.
Also, $x+y+z=28$,
Subtracting (2) from (1) gives
$$
(b-1)(b x+x+y)=1971 \text {. }
$$

So, $(b-1) \mid 1971=9 \times 3 \times 73$.
$$
\begin{array}{l}
\because 12<b<45, \\
\therefore b-1=3 \times 9 . \\
\therefore b=28 .
\end{array}
$$
[Solution 2] (Elimination method)
$$
\begin{array}{l}
(1999)_{10}=(1558)_{11}, \\
(1959)_{10}=(8 \overline{134}) 15,(x=3, y=13, z=4) \\
(1999)_{10}=(631)_{12} . \\
(1990)_{10}=(2 \overline{15} \overline{11})_{28} .(x=2, y=15, z=11)
\end{array}
$$