27. The number of positive integers $n$ that make $3^{n}+81$ a perfect square is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
27. B.

When $n \leqslant 4$, it is easy to see that $3^{n}+81$ is not a perfect square. Therefore, assume $n=k+4$, where $k$ is a positive integer. Then
$$
3^{k}+81=81\left(3^{k}+1\right) \text {. }
$$

Since 81 is a perfect square, there must exist a positive integer $x$ such that
$$
3^{k}+1=x^{2} \Leftrightarrow(x-1)(x+1)=x^{2}-1=3^{k} \text {. }
$$

Therefore, there must exist $k$ such that $x-1$ and $x+1$ are both powers of 3.
Since two numbers differ by 2 and are both powers of 3, they can only be 1 and 3, thus $x=2, k=1$. Therefore, $\eta=\dot{k}+4=5$ is the only positive integer that makes $3^{r}+81$ a perfect square.