219. Three men have a certain sum of money, the part of the first being half, the second a third, and the third a sixth of the entire sum.

Wishing to save part of the money, each takes from the total sum as much as he can carry, after which the first deposits half, the second a third, and the third a sixth of what each carried. After some time, they take these funds back, and each has to receive $\frac{1}{3}$ of the entire sum that was on deposit. How much money did each have?

Problems on systems of equations from "Liber abaci".
219. John of Palermo, master, court mathematician, who proposed this problem to Leonardo Fibonacci at a tournament.

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-130.jpg?height=58&width=1282&top_left_y=1590&top_left_x=427)
Then we have the system:

$$
\begin{aligned}
& \frac{t}{2}=\frac{x}{2}+\frac{1}{3}\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right) \\
& \frac{t}{3}=\frac{2}{3} y+\frac{1}{3}\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right) \\
& \frac{t}{6}={ }_{6}^{5} z+\frac{1}{3}\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}\right)
\end{aligned}
$$

Solving this system, we find the smallest value \( y=13 \) and \( t=47 \), from which \( z=1 \) and \( x=33 \). The share of the first is \( 23 \frac{1}{2} \), the second \( 15 \frac{2}{3} \), and the third \( 7 \frac{5}{6} \). The problem has three solutions.

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