18. Let $k \in \mathbf{Z}_{+}$. If $\frac{7}{51}$ is written in base $k$ as
$$
\frac{7}{51}=\overline{0.23}_{k}=(0.2323)_{k} \text {, }
$$

then $k=$ ( ).
(A) 13
(B) 14
(C) 15
(D) 16
(E) 17
18. D.

From the problem, we know
$$
\begin{aligned}
k^{2} \overline{0.23}_{k} & =(23.2323 \cdots)_{k} \\
& \Rightarrow\left(k^{2}-1\right) \overline{0.23}_{k}=(23)_{k}=2 k+3 \\
& \Rightarrow \frac{7}{51}=\overline{0.23_{k}}=\frac{2 k+3}{k^{2}-1} \\
& \Rightarrow 7 k^{2}-102 k-160=0 \\
& \Rightarrow(7 k+10)(k-16)=0 \\
& \Rightarrow k=16 .
\end{aligned}
$$