2. (5 points) Given: $1-\frac{1}{6+\frac{1}{6+\frac{1}{6}}}=\frac{1}{\mathrm{~A}+\frac{1}{\mathrm{~B}+\frac{1}{1}}}$ where $A$, $B$, and $C$ are all natural numbers greater than 0 and distinct from each other,
then $(A+B) \div C=$

Translate the above text into English, preserving the original text's line breaks and format, and output the translation result directly.
【Solution】Solution: $1-\frac{1}{6+\frac{1}{6+\frac{1}{6}}}$,
$$
\begin{array}{l}
=1-\frac{1}{6+\frac{6}{37}}, \\
=1-\frac{1}{\frac{228}{37}}, \\
=1-\frac{37}{228}, \\
=\frac{191}{228} .
\end{array}
$$

So $\frac{1}{B+\frac{1}{B+\frac{1}{C}}}=\frac{191}{228}$,
that is, $\mathrm{A}+\frac{1}{\mathrm{~B}+\frac{1}{\mathrm{C}+\frac{1}{\mathrm{C}}}}=\frac{228}{191}<2$,
so $A$ can only be 1;
then $\frac{1}{\mathrm{~B}+\frac{1}{\mathrm{C}+\frac{1}{\mathrm{C}}}}=\frac{228}{191}-1=\frac{37}{191}$,
that is: $\mathrm{B}+\frac{1}{\mathrm{C}+\frac{1}{\mathrm{C}}}=\frac{191}{37}=5 \frac{6}{37}$,
so $B$ can only be $1, 2, 3, 4, 5$.
After enumeration, only 5 is suitable;
then $\frac{1}{\mathrm{C}+\frac{1}{\mathrm{C}}}=5 \frac{6}{37}-5=\frac{6}{37}$,
so $C=6$.
Therefore: $(A+B) \div C$,
$=(1+5) \div 6$,
$=1$.
The answer is: 1.