Karel, Mirek, and Luděk were comparing their stamp collections. When they checked the numbers, they found that Karel and Mirek had 101 stamps together, Karel and Luděk 115 stamps, and Mirek and Luděk 110. When they verified what they could exchange, they found that none of them had the same stamp, but Karel and Mirek had 5 stamps in common, Karel and Luděk 12 in common, and Mirek and Luděk 7.

How many stamps does Luděk have that are different from the other boys?

(M. Smitková)
Let's denote the number of stamps owned by Karel, Mirek, and Luděk as $K$, $L$, and $M$ respectively. According to the problem, we have:

$$
K+M=101, \quad K+L=115, \quad M+L=110
$$

By summing these three equations and performing further manipulations, we get:

$$
\begin{aligned}
2 K+2 M+2 L & =326, \\
K+M+L & =163 \\
L & =163-(K+M) .
\end{aligned}
$$

Substituting the first equation from the initial set, we find that Luděk has a total of 62 stamps $(L=163-101=62)$.

Out of these 62 stamps, 12 are the same as Karel's and 7 are the same as Mirek's. Therefore, Luděk has 43 stamps that are different from those of the other boys $(62-12-7=43)$.

Another solution. Let's denote the number of stamps owned exclusively by Karel, Mirek, and Luděk as $k$, $l$, and $m$ respectively. According to the problem, we have:

$$
\begin{aligned}
(k+5+12)+(m+5+7) & =101, \\
(k+5+12)+(l+7+12) & =115, \\
(m+5+7)+(l+7+12) & =110 .
\end{aligned}
$$

By simplifying each line, we get the equivalent system:

$$
\begin{aligned}
k+m & =72, \\
k+l & =79 \\
m+l & =79 .
\end{aligned}
$$

From the last two equations, it follows that $k=m$, and from the first equation, we get $k=m=36$. From this, we can further calculate $l=43$. Therefore, Luděk has 43 stamps that are different from those of the other boys.

Note. The information about the number of stamps can be visualized using a Venn diagram as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_e6d56f2031023e9db813g-4.jpg?height=454&width=714&top_left_y=818&top_left_x=677)

When expressing the sums of the stamps, the values from the relevant intersections are counted twice. This insight can be more clearly illustrated as follows:
![](https://cdn.mathpix.com/cropped/2024_04_17_e6d56f2031023e9db813g-4.jpg?height=640&width=938&top_left_y=1492&top_left_x=567)

Evaluation. 3 points for expressing the information from the problem using equations and their manipulation; 3 points for solving the system and drawing the conclusion.