A cycle of three conferences had constant success, that is, in each session there was the same number of participants. However, half of those who attended the first one did not return; one-third of those who attended the second conference attended only that one, and one-quarter of those who attended the third did not attend either the first or the second. Knowing that there were 300 registered and that each one attended at least one conference, determine:

a) How many people attended each conference?

b) How many people attended all three conferences?

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Solution

a) Let's call $P$ the number of gifts at each conference, $x, y, t$ and $z$ will be the integers representing those who attended all three conferences; only the first and second; only the first and third; and only the second and third, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-20.jpg?height=668&width=621&top_left_y=586&top_left_x=625)

We can then write that:

- $y=300-\left(P+\frac{P}{2}+\frac{P}{3}\right)=300-\frac{11 P}{6}$;
- $z=300-\left(P+\frac{P}{3}+\frac{P}{4}\right)=300-\frac{19 P}{12}$;
- $t=300-\left(P+\frac{P}{2}+\frac{P}{4}\right)=300-\frac{7 P}{4}$;
- $x=\frac{P}{2}-(y+t)=\frac{49 P}{12}-600$.

Since $x$ is an integer, $P$ is a multiple of 12, so there exists an integer $k$ such that $P=12 k$. Thus, we have $y=300-22 k, z=300-19 k, t=300-21 k$ and $x=49 k-600$, all non-negative, hence:

- $y=300-22 k \geq 0 \Rightarrow k \leq 13.63$,
- $z=300-19 k \geq 0 \Rightarrow k \leq 15.78$,
- $t=300-21 k \geq 0 \Rightarrow k \leq 14.28$,
- $x=49 k-600 \geq 0 \Rightarrow k \geq 12.24$.

The only integer in this interval is $k=13$. This produces $P=156$.

b) From the previous item, $x=49 k-600=49 \cdot 13-600=37$.