At the end of a day of activities, an amusement park collected 100 reais from the tickets of 100 people. We know that each adult had to pay 3 reais to enter, each young person 2 reais, and each child 30 cents. What is the minimum number of adults who entered the park on that day?

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Solution

To find the minimum number of adults, we will divide the problem into cases:

a) Suppose it is possible to have 0 adults. Thus, the sum of the quantities of young people and children is 100. Consequently, if each child pays 1.70 more to match the contribution of the young people, we will have $2 \cdot 100=200$ reais. Since the actual contribution of these people was 100, the difference $200-100=100$ corresponds to the extra contribution of the children. Therefore, the number of children is $\frac{100}{1.7}$, which is not an integer. This shows that it is not possible to have 0 adults.

b) Suppose it is possible to have 1 adult. Thus, the sum of the quantities of young people and children is 99. Again, if each child pays 1.70 more to match the contribution of the young people, we will have $2 \cdot 99=198$ reais. Since the actual contribution of these people was $100-3 \cdot 1=97$, the difference $198-97=101$ corresponds to the extra contribution of the children. Therefore, the number of children is $\frac{101}{1.7}$, which is not an integer. This shows that it is not possible to have only 1 adult.

c) To show that the minimum number of adults required is 2, it suffices to exhibit compatible quantities of young people and children with the data from the problem. Repeating the strategy from the previous cases of simulating the extra payment of 1.70 by each child, we can conclude that the number of children needed with 2 adults is

$$
\frac{2 \cdot (100-2)-(100-2 \cdot 3)}{1.7}=60
$$

Thus, with 2 adults, the quantities of young people and children are $100-2-60=38$ and 60, respectively. Note that $3 \cdot 2+2 \cdot 38+0.3 \cdot 60=100$.

Therefore, at least 2 adults are necessary.

Observation: Assuming that $x, y$, and $z$ represent the quantities of adults, young people, and children, respectively, we can obtain another solution by writing:

$$
x+y+z=100 \text{ and } 3x+2y+0.3z=100
$$

Then, it suffices to multiply the first equation by 20 and subtract it from the second equation multiplied by 10, obtaining $10x-17z=-1000$. From this, 17 divides $10(x+100)$. Since 100 leaves a remainder of 15 when divided by 17 and $\text{gcd}(10,17)=1$, $x$ must leave a remainder of 2 when divided by 17. The smallest non-negative integer that leaves a remainder of 2 when divided by 17 is 2. It then suffices to follow the final steps of the previous solution to exhibit an example with 2 adults.