Find three numbers such that if each gives a certain specified part of itself to the next, they will be equal after the mutual gift-giving. For example, the first number gives one-third of its value to the second, the second gives one-fourth of its value to the third, and finally, the third gives one-fifth of its value to the first. Determine the sequence of positive integer solutions.
Let the three numbers be of the form $3 x, 4 y, 5 z$; after mutual gift-giving, they become:

$$
2 x+z ; 3 y+x ; 4 z+y
$$

which must have equal differences. This condition yields two equations with three unknowns:

$$
\begin{aligned}
& 2 x+z=3 y+x \\
& 3 y+x=4 z+y
\end{aligned}
$$

From the second equation, $z=\frac{2 y+x}{4}$ can only be an integer if $2 y+x=4 k$, where $k$ is any integer; $k$ can express the other unknowns, and indeed

$$
z=k ; y=k ; x=2 k
$$

thus, the three sought numbers can be expressed in the form $6 k, 4 k, 5 k$.