4.061. Solve the equation

$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x}+\ldots+\frac{1}{x}=3
$$

where $x$ is a positive integer.
## Solution.

Multiplying both sides of the equation by $x$, we have

$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$

The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1, a_{n}=1$,

$$
n=\frac{a_{n}-a_{1}}{d}+1=\frac{1-x+1}{-1}+1=x-1 . \text { Since } S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n
$$

then $\frac{x-1+1}{2} \cdot(x-1)=3 x, x^{2}=7 x, x=7(x \neq 0)$.

Answer: $x=7$.