6. Problem: How many pairs of positive integers $(a, b)$ with $\leq b$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{1}{6}$ ?
Solution: $\frac{1}{a}+\frac{1}{b}=\frac{1}{6} \Rightarrow \frac{a+b}{a b}=\frac{1}{6} \Rightarrow a b=6 a+6 b \Rightarrow a b-6 a-6 b=0$. Factoring yields $(a-b)(b-6)-36=0$. Then $(a-6)(b-6)=36$. Because $a$ and $b$ are positive integers, only the factor pairs of 36 are possible values of $a-6$ and $b-6$. The possible pairs are:
$$
\begin{array}{l}
a-6=1, b-6=36 \\
a-6=2, b-6=18 \\
a-6=3, b-6=12 \\
a-6=4, b-6=9 \\
a-6=6, b-6=6
\end{array}
$$

Because $a \leq b$, the symmetric cases, such as $a-6=12, b-6=3$ are not applicable. Then there are 5 possible pairs.