$29 \cdot 65$ Find the smallest positive value of $n$ for which $\frac{n-13}{5 n+6}$ is a non-zero reducible fraction.
(A) 45.
(B) 68.
(C) 155.
(D) 226.
(E) None of the above.
(36th American High School Mathematics Examination, 1985)
[Solution 1] From the problem, if $n-13$ and $5n+6$ have a common divisor, then they are both multiples of some positive number $k (k \neq 1)$.
$\therefore$ Their difference should also be a multiple of $k$, and since $5 \nmid (5n+6)$, then
$$
(5n+6) - 5(n-13)
$$

should also be a multiple of $k$.
And $5n+6 - 5(n-13) = 71$, and 71 is a prime number, so $k=71$.
$$
\text{Also, } 71 \mid (n-13),
$$

thus $n = 71m + 13, \quad m = 0, 1, 2, \cdots$
$\because \frac{n-13}{5n+6}$ is not $0, \therefore m=1$ when $n=84$ is the solution.
Therefore, the answer is $(E)$.
[Solution 2] Let $n-13 = ms, 5n+6 = mt (m \neq 1, s, t$ are integers).
Solving, we get
$$
m = \frac{71}{t-5s}, \quad n = \frac{13t + 6s}{t-5s}.
$$
$\because \quad 71$ is a prime number, and $m$ is a positive integer, then $m=71$.
$\therefore n = 71s + 13, s$ is a positive integer.
When $s=1$, $n=84$.
Therefore, the answer is $(E)$.