117. Find all integers $k$ such that the quadratic equation $k x^{2}-2 \times(3 k-1) x+9 k-1=0$ has at least one integer root.
Solution: From the original equation, we get
$$
k(x-3)^{2}=1-2 x \text {. }
$$
$x$ is an integer root, clearly $x \neq 3, k \neq 0$. Thus, $k=$ $\frac{1-2 x}{(x-3)^{2}}$. Furthermore, since $k$ is an integer, we have
$$
|1-2 x| \geqslant(x-3)^{2} \text {. }
$$

If $x \leqslant 0$, then $1-2 x \geqslant x^{2}-6 x+9$, which simplifies to $(x-2)^{2}+4 \leqslant$ 0, which is impossible.

If $x>0$, then $2 x-1 \geqslant x^{2}-6 x+9$, which simplifies to $(x-4)^{2} \leqslant 6$, or $|x-4| \leqslant \sqrt{6}$. Thus, $|x-4| \leqslant 2$. Therefore, $x$ can only be 2, 4, 5, 6.

Upon individual verification, it is found that only $x=2$ (corresponding to $k=-3$) and $x=4$ (corresponding to $k=-7$) satisfy the conditions.
In summary, the required values of $k$ are -3 or -7.
(Wu Weizhao, Department of Mathematics, College of Science, Guangzhou University, 510405)