Problem 3.6. In a meeting between four UN countries, let's say $A, B, C$, and $D$, country $A$ has twice as many representatives as $B$, three times as many as $C$, and four times as many as $D$. It is intended to distribute the representatives at tables with the same number of people at each one. There is only one condition: in each table, any of the countries must be in a numerical minority compared to the other three combined. How many representatives should there be at each table, at a minimum?
Solution to Problem 3.6 The answer is 25. Let's see the proof. Let $a, b, c$ and $d$ be the number of representatives from each country. Since $a$ must be a multiple of 3 and 4, it must also be a multiple of 12. Therefore, there exists a number $k$ such that $a=12k$, then $b=6k, c=4k$ and $d=3k$. The total number of representatives is then $25k$.

If we call $M$ the number of tables, and $P$ the number of people at each table, we have $MP=25k$.

Let $a_{i}$ be the number of representatives from country $A$ at table number $i$. The condition imposed tells us that $a_{i}<\frac{P}{2}$, or $2a_{i}<P$. Since $a_{i}$ is an integer, this implies that $2a_{i} \leq P-1$. Adding all the $a_{i}$, we get:

$$
24k=2a=2a_{1}+\cdots+2a_{M} \leq M(P-1)=25k-M
$$

From this, we deduce $M \leq k$. Therefore, since $MP=25k$, we finally deduce that $P \geq 25$.

It only remains to show that, indeed, a configuration can be achieved where there are 25 people at each table. This can be achieved, for example, with a single table where there are 12 representatives from country $A$, 6 from $B$, 4 from $C$, and 3 from $D$.