35. (1985 American Mathematical Invitation Contest) Let the four vertices of a regular tetrahedron be $A, B, C, D$, with each edge length being 1 meter. A small insect starts from point $A$ and moves according to the following rule: at each vertex, it chooses one of the three edges connected to that vertex with equal probability, and crawls all the way to the end of the chosen edge. Suppose the probability that the insect is exactly at vertex $A$ after crawling 7 meters is $p=\frac{n}{729}$. Find the value of $n$.
35. Let $a_{n}$ denote the probability that the bug returns to point $A$ after walking $n$ meters $(n=0,1)$. If the bug has walked $n-1$ meters and is not at $A$, the probability is $1-a_{n-1}$, and the probability of moving from a point outside $A$ to $A$ is $\frac{1}{3}$. Therefore, we have
$$
a_{n}=\frac{1}{3}\left(1-a_{n-1}\right) .
$$

Given $a_{0}=1$, meaning the bug starts at $A$, we can calculate step by step from (1):
$$
\begin{array}{l}
a_{1}=0, a_{2}=\frac{1}{3}, a_{3}=\frac{2}{9}, a_{4}=\frac{7}{27}, \\
a_{5}=\frac{20}{81}, a_{6}=\frac{61}{243}, a_{7}=\frac{182}{729} .
\end{array}
$$

From $a_{7}=\frac{n}{729}=\frac{182}{729}$, we get $n=182$.
Alternative solution: We can also find the general formula for $a_{n}$ from (1): $a_{n}-\frac{1}{4}=-\frac{1}{3}\left(a_{n-1}-\frac{1}{4}\right)$, which leads to $a_{n}-\frac{1}{4}=\left(-\frac{1}{3}\right)^{n}\left(a_{0}-\frac{1}{4}\right)$, or $a_{n}-\frac{1}{4}=\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n}$,

Thus, $a_{n}=\frac{3}{4} \cdot\left(-\frac{1}{3}\right)^{n}+\frac{1}{4}(n=0,1,2, \cdots)$.
That is, $a_{n}=\frac{1}{4}+\frac{3}{4}\left(-\frac{1}{3}\right)^{n}$, and we calculate $a_{7}=\frac{182}{729}$.
Therefore, $n=182$.