27. Consider a sequence $x_{1}, x_{2}, x_{3} \cdots$, where $x_{1}=\sqrt[3]{3}, x_{2}=(\sqrt[3]{3}) \sqrt[3]{3},$ and in general, $x_{n}=\left(x_{n-1}\right) \sqrt[3]{{ }^{3}}$. For $n>1$, what is the smallest $n$ such that $x_{n}$ is an integer?
(A) 2 ;
(B)
3; (C) 4;
D) 9 ; (E) 27.
27. We $\quad \therefore=3^{\frac{1}{3}}$
$$
\begin{array}{l}
x_{4}=(3 \sqrt[3]{9 / 3}) \sqrt[3]{3}=3^{\sqrt[3]{27 / 3}}=3^{3 / 3} \\
=3 \text {. } \\
\end{array}
$$

We also need to verify that $x_{1}, x_{2}, x_{3}$ are not integers. It is easy to see that $x_{1}\sqrt{2}$, and we need to prove that $x_{3}>2$. Therefore, the smallest $n$ value that makes $x_{n}$ an integer is 4.
(C) True,