In a small kingdom, the subjects came to greet the new king and his minister. Each brought some gift: the 60 poorest brought only a handmade wooden figurine, the richer ones either 2 gold coins or 3 silver coins. The number of silver coins given was more than 40 but less than 100. The minister received all the figurines and also one-seventh of all the gold coins and one-third of all the silver coins. The king and his minister received the same number of items.

Determine how many subjects there could have been, how many gold coins could have been given, and how many silver coins. 

(M. Volfová)
Let the number of gold coins be $z$ and the number of silver coins be $s$. The gold coins were given in pairs, so $z$ is an even number, and the silver coins were given in threes, so $s$ is divisible by three. Additionally, $s$ is between 40 and 100. The minister received 60 figurines and $\frac{1}{7} z + \frac{1}{3} s$ coins, while the king received $\frac{6}{7} z + \frac{2}{3} s$ coins. This implies that $z$ must be divisible by 7, which, combined with the previous condition, means that $z$ must be divisible by 14. Both the king and his minister received the same number of items, which in our notation gives the equation

$$
60 + \frac{z}{7} + \frac{s}{3} = \frac{6z}{7} + \frac{2s}{3}
$$

After simplification,

$$
\begin{aligned}
\frac{5z}{7} + \frac{s}{3} & = 60 \\
15z + 7s & = 7 \cdot 3 \cdot 60.
\end{aligned}
$$

We are looking for all integer solutions to equation (2) that satisfy the conditions mentioned above. If we express the unknown $z$,

$$
z = 84 - \frac{7s}{15}
$$

we see that $s$ must be divisible by 15. Since $z$ must be even, the value of the expression $\frac{7}{15} s$ must also be even, which means that $s$ must be an even number. Therefore, $s$ must be divisible by 30, which, considering the condition $40 < s < 100$, means that $s$ can only be 60 and 90. For both these possibilities, we can calculate $z$ using (3), and the number of subjects can be determined as $60 + \frac{1}{3} s + \frac{1}{2} z$. The problem has the following two solutions (note that all the conditions mentioned above are satisfied):

| silver coins | gold coins | subjects |
| :---: | :---: | :---: |
| 60 | 56 | 108 |
| 90 | 42 | 111 |

Another solution. Equation (2) can also be solved by expressing the unknown $s$:

$$
s = 180 - \frac{15z}{7}.
$$

From the beginning, we know that $z$ must be divisible by 14, so we substitute multiples of 14 for $z$ and check if $40 < s < 100$. If so, we calculate the number of subjects as $60 + \frac{1}{3} s + \frac{1}{2} z$ (note that the other conditions mentioned above are satisfied):

| gold coins | silver coins | subjects |
| :---: | :---: | :---: |
| 14 | 150 |  |
| 28 | 120 |  |
| $\mathbf{42}$ | $\mathbf{90}$ | $\mathbf{111}$ |
| $\mathbf{56}$ | $\mathbf{60}$ | $\mathbf{108}$ |
| 70 | 30 |  |

We do not continue further because $s$ is less than 40; the problem has two solutions.

Evaluation. 2 points for setting up equation (1) and its subsequent simplification; 2 points for finding one possibility; 1 point for finding the second possibility; 1 point for justifying that there are no more solutions.