Pan Stříbrný uspořádal výstavu. Vystavoval 120 prstenů, které ležely na stolech podél stěn sálu a tvořily tak jednu velkou kružnici. Prohlídka začínala u vchodových dveří v označeném směru. Odtud každý třetí prsten v řadě byl zlatý, každý čtvrtý prsten byl starožitný a každý desátý prsten měl diamant. Prsten, který neměl žádnou z těchto tří vlastností, byl padělek.

Kolik bylo na výstavě zlatých prstenů, které byly starožitné a zároveň měly diamant? Kolik vystavil pan Stříbrný padělků?

(L. Hozová)

Nápověda. Podle jakých pravidel byly rozmístěny prsteny s různými kombinacemi třech zmiňovaných vlastností?

Mr. Silver organized an exhibition. He displayed 120 rings, which were placed on tables along the walls of the hall, forming one large circle. The tour started at the entrance door in a marked direction. From there, every third ring in the row was gold, every fourth ring was antique, and every tenth ring had a diamond. A ring that did not have any of these three properties was a counterfeit.

How many gold rings were on display that were also antique and had a diamond? How many counterfeits did Mr. Silver display?

(L. Hozová)

Hint. According to what rules were the rings with different combinations of the three mentioned properties placed?
Every 3rd ring was golden, every 4th was antique, and every 10th had a diamond. Thus,

- the number of golden rings was $120: 3=40$,
- the number of antique rings was $120: 4=30$,
- the number of rings with a diamond was $120: 10=12$.

When counting rings with multiple properties, we first determine the regularity with which they appeared at the exhibition: every 12th ring was both golden and antique, every 30th ring was golden with a diamond, and every 20th ring was antique with a diamond (here, 20 is the least common multiple of the numbers 4 and 10). Thus,

- the number of golden antique rings was $120: 12=10$,
- the number of golden rings with a diamond was $120: 30=4$,
- the number of antique rings with a diamond was $120: 20=6$.

Furthermore, every 60th ring had all three properties (60 is the least common multiple of the numbers 3, 4, and 10), thus

- the number of golden antique rings with a diamond was $120: 60=2$.

When counting rings with any of the mentioned properties, we must be cautious: 10 rings were both golden and antique, 2 of which also had a diamond, so the number of golden antique rings without a diamond was $10-2=8$. Similarly, the number of golden non-antique rings with a diamond was $4-2=2$ and the number of non-golden antique rings with a diamond was $6-2=4$.

The number of golden rings with some additional property was $2+8+2=12$, while the total number of golden rings was 40, so the number of golden non-antique rings without a diamond was $40-12=28$. Similarly, the number of non-golden antique rings without a diamond was $30-(2+8+4)=16$ and the number of non-golden non-antique rings with a diamond was $12-(2+2+4)=4$.

The previous counts and relationships can be illustrated using a Venn diagram as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_fc767241fb97c8703187g-4.jpg?height=448&width=780&top_left_y=2112&top_left_x=638)

The number of rings with any of the three properties (i.e., rings that were not fakes) was $2+8+4+2+28+16+4=64$. Therefore, the number of fakes was $120-64=56$.

Note. If we denote the three basic sets of rings as $Z, S$, and $D$, the initial part of the solution can be summarized as follows:

$$
\begin{gathered}
|Z|=40, \quad|S|=30, \quad|D|=12 \\
|Z \cap S|=10, \quad|Z \cap D|=4, \quad|S \cap D|=6, \\
|Z \cap S \cap D|=2 .
\end{gathered}
$$

In the next part, we determined the number of elements in the union $Z \cup S \cup D$ by gradually expressing the number of elements in mutually disjoint* subsets $(Z \cap S) \backslash(Z \cap S \cap D)$, $(Z \cap D) \backslash(Z \cap S \cap D)$, etc., which we then summed. More concisely, the same result can be obtained by the following calculation:

$$
\begin{aligned}
|Z \cup S \cup D| & =|Z|+|S|+|D|-|Z \cap S|-|Z \cap D|-|S \cap D|+|Z \cap S \cap D|= \\
& =40+30+12-10-4-6+2=64 .
\end{aligned}
$$

This relationship is known as the principle of inclusion and exclusion. To justify it generally (or to further generalize it), it suffices to verify that each of the disjoint parts of the Venn diagram is counted exactly once.