263 Given that 2009 can be written as a three-digit number $x y z$ in base $k$. If $x+y+z=20+0+9$. Try to determine $x 、 y 、 z 、 k$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Solve: From the given conditions, we have
$$
\left\{\begin{array}{l}
x k^{2}+y k+z=2009, \\
x+y+z=29 .
\end{array}\right.
$$
(1) - (2) gives
$$
(k-1)[(k+1) x+y]=1980 \text {. }
$$

From the problem, we know $k>12\left(12^{3}=17282009\right)$.
Therefore, $k-1>11$.
Since $k-1<(k+1) x+y, 1980=44 \times 45$,
Thus, $k-1<44$.
Hence, $11<k-1<44$.
From equation (3), we get $(k-1) \mid 1980$.
Also, $1980=2 \times 2 \times 3 \times 3 \times 5 \times 11$, so among all the positive divisors of 1980, those greater than 11 and less than 44 are $12, 15, 18, 20, 22, 30, 33$. Therefore, $k$ can take the values 13, 16, 19, 21, 23, 31, 34.
(1) When $k=13$, $2009=11 \times 13^{2}+11 \times 13+7$;
(2) When $k=16$, $2009=7 \times 16^{2}+13 \times 16+9$;
(3) When $k=19$, $2009=5 \times 19^{2}+10 \times 19+14$;
(4) When $k=21$, $2009=4 \times 21^{2}+11 \times 21+14$;
(5) When $k=23$, $2009=3 \times 23^{2}+18 \times 23+8$;
(6) When $k=31$, $2009=2 \times 31^{2}+2 \times 31+25$;
(7) When $k=34$, $2009=1 \times 34^{2}+25 \times 34+3$.
In summary, we get 7 sets of solutions:
$$
\begin{array}{l}
(x, y \vee z \vee k)=(11,11,7,13),(7,13,9,16), \\
(5,10,14,19),(4,11,14,21),(3,18,8,23), \\
(2,2,25,31),(1,25,3,34) .
\end{array}
$$