Problem 3. Determine the largest natural number $n$ for which the following condition is satisfied:

there exist $n$ distinct half-lines, two by two, with the same origin, such that the measure of any angle formed by these half-lines is a natural number that is not prime.
Solution. If we take 90 angles of $4^{\circ}$, formed around a point, we obtain 90 rays that form angles with integer measures, multiples of $4 \ldots \ldots . .2$ p

Let's show that if we take 91 or more rays, the condition is not satisfied. We denote the angles formed around the common origin of the rays, in the direction of the clock hands, starting from any angle, as $u_{1}, u_{2}, \ldots, u_{n}, n \geq 91$. Then the sums $u_{1}+u_{2}, u_{3}+u_{4}, \ldots, u_{89}+u_{90}$ together have at most $359^{\circ}$. Since there are 45 sums and $8 \cdot 45=360$, there exists a sum $S<8^{\circ}$. Since $S \geq 2$, it follows that $S$ is a prime number or $S=4=2+2$, or $S=6=2+4=3+3$; in all cases, we deduce that $S$ or one of its terms is a prime number.

Therefore, the required number is 90 $1 p$