Pankrác, Servác, and Bonifác are brothers who are $P, S$, and $B$ years old. We know that $P, S$, and $B$ are natural numbers less than 16, for which the following equations hold:

$$
\begin{aligned}
P & =\frac{5}{2}(B-S), \\
S & =2(B-P), \\
B & =8(S-P) .
\end{aligned}
$$

Determine the age of all three brothers.

(L. Hozová)

Hint. Bonifác's age can be determined very easily.
From the third equation, it follows that $B$ is a natural number less than 16 if and only if $S-P=1$, or $S=P+1$; then necessarily $B=8$. Substituting these findings into the second equation, we determine $P$:

$$
\begin{aligned}
P+1 & =2(8-P) \\
P+1 & =16-2 P \\
3 P & =15 \\
P & =5 .
\end{aligned}
$$

Thus, $S=5+1=6$ and it is easy to verify that the triplet $B=8, P=5$, and $S=6$ also satisfies the first equation: $5=\frac{5}{2}(8-6)$. Therefore, Pankrác is 5 years old, Servác is 6 years old, and Bonifác is 8 years old.

Note. The equations from the problem can be manipulated in various ways, however, without the restriction $P, S, B > P > 0$. Just like in the previous solution, we determine from the third equation that $B=8$ and $P=S-1$. Additionally, it is clear from the second equation that $S$ is an even number. Therefore, we see that the solution to the problem can only be one of the following triplets of numbers:

| $B$ | $S$ | $P$ |
| :---: | :---: | :---: |
| 8 | 6 | 5 |
| 8 | 4 | 3 |
| 8 | 2 | 1 |

By substituting into the first and second equations, we find that the only solution is the triplet $B=8, S=6$, and $P=5$.