In 1988, the Chinese Junior High School Mathematics League had the following problem: If natural numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=x_{1} x_{2} x_{3} x_{4} x_{5}$, what is the maximum value of $x_{5}$?
This problem is novel and unique, and some students find it difficult to start. Below, I will first discuss the solution to this problem.

Algorithm 1: Given the symmetry of the equation, we can assume without loss of generality that \( x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant x_{5} \), and then use the "bounding method":
\[
\begin{array}{l}
1=\frac{1}{x_{1} x_{2} x_{3} x_{4}}+\frac{1}{x_{1} x_{2} x_{3} x_{5}}+\frac{1}{x_{1} x_{3} x_{4} x_{5}} \\
+\frac{1}{x_{2} x_{3} x_{4} x_{5}} \leqslant \frac{1}{x_{4}}+\frac{1}{x_{5}}+\frac{1}{x_{4} x_{5}}+\frac{1}{x_{4} x_{5}} \\
+\frac{1}{x_{4} x_{5}}=\frac{3+x_{4}+x_{5}}{x_{4} x_{5}}
\end{array}
\]

Therefore,
\[
\begin{array}{l}
x_{4} x_{3} \leqslant 3+x_{4}+x_{5}, \\
\left(x_{4}-1\right)\left(x_{5}-1\right) \leqslant 4 .
\end{array}
\]

If \( x_{4} = 1 \), then \( x_{1} = x_{2} = x_{3} = x_{4} = 1 \), and the original equation becomes \( 4 + x_{5} = x_{5} \), which is a contradiction.
If \( x_{4} > 1 \), then \( x_{5} - 1 \leqslant 4 \), i.e., \( x_{5} \leqslant 5 \).
Thus, the maximum value of \( x_{5} \) is 5. It is easy to observe that one solution is \( (1,1,1,2,5) \).

Solution 2: When \( x_{5} \) is maximized, it should be no less than any \( x_{i} \) (for \( i=1,2,3,4 \)), so
\[
\begin{array}{l}
x_{5} < x_{2} + x_{2} + x_{3} + x_{4} + x_{5} \leqslant 5 x_{3}^{3}, \\
x_{3} < x_{1} x_{2} x_{3} x_{4} x_{5} \leqslant 5 x_{3}, \\
1 < x_{1} x_{2} x_{3} x_{4} \leqslant 5 .
\end{array}
\]

From the original equation, we get \( x_{5} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{x_{1} x_{2} x_{3} x_{4} - 1} \).
If \( x_{1} x_{2} x_{3} x_{4} = 2 \), then among \( x_{1}, x_{2}, x_{3}, x_{4} \), there is exactly one 2 and three 1s. Substituting, we get \( x_{5} = 5 \).

For \( x_{1} x_{2} x_{3} x_{4} = 3, 4, 5 \), similarly, \( x_{5} = 3 \) and \( x_{5} = 2 \).