1. In the Rhind Papyrus (Ancient Egypt), among other information, there are decompositions of fractions into the sum of fractions with a numerator of 1, for example,

$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$

One of the denominators here is replaced by the letter $x$. Find this denominator.
# Solution:

First, find $\stackrel{1}{-.}$ from the equation:

$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$

$\frac{2}{73}-\frac{1}{219}-\frac{1}{292}-\frac{1}{60}=\frac{1}{x}$

$\frac{2}{73}-\frac{1}{73 \cdot 3}-\frac{1}{73 \cdot 4}-\frac{1}{60}=\frac{1}{x}$

$\frac{2 \cdot 3 \cdot 4-4-3}{73 \cdot 3 \cdot 4}-\frac{1}{60}=\frac{1}{x}$
$\frac{17}{73 \cdot 3 \cdot 4}-\frac{1}{5 \cdot 3 \cdot 4}=\frac{1}{x}$

$\frac{17 \cdot 5-73}{73 \cdot 3 \cdot 4 \cdot 5}=\frac{1}{x}$

$\frac{12}{73 \cdot 3 \cdot 4 \cdot 5}=\frac{1}{x}$

$\frac{1}{73 \cdot 5}=\frac{1}{x}$

$\frac{1}{365}=\frac{1}{x}$

Thus, $x=365$.

Answer: $x=365$.