## Solution to Exercise 11

We can try to color the numbers greedily: 1 can be colored with a color we denote as $a$, 2 and 3 with the same color $b$ (but not color $a$), then 4 and 6 with color $c$ (we can also color 5 and 7 with color $c$), etc. It seems that an efficient coloring is to color all numbers $n$ such that $2^k \leq n < 2^{k+1}$ with color $k+1$, as long as $n \leq 2023$. Thus, we color $[1,2[$ with color 1, $[2,4[$ with color 2, ..., $[1024,2023]$ with color 11. If $a \neq b$ are of the same color $k \in \{1, \ldots, 11\}$, then $0 < \frac{a}{b} < \frac{2^{k+1}}{2^k} = 2$, so if $b$ divides $a$, then $\frac{a}{b}$ is an integer and must be 1. Thus, $a = b$, which is a contradiction. Therefore, our coloring satisfies the condition of the problem and requires 11 colors.

Conversely, if a coloring satisfies the problem statement, the numbers $2^0 = 1, 2^1, \ldots, 2^{10}$ are between 1 and 2023, and if we take $a \neq b$ among these 11 numbers, either $a$ divides $b$ or $b$ divides $a$. Thus, these 11 numbers must be of different colors. Therefore, at least 11 colors are needed.

Thus, the minimum number of colors required is 11.

Comment from the graders: The exercise is very well done. As in any problem where we need to find the minimum number such that, it is important to split the proof into two parts: why $n=11$ colors are sufficient, and why if $n \leq 10$, $n$ colors are not sufficient.
Exercise 11. Félix wishes to color the integers from 1 to 2023 such that if $a, b$ are two distinct integers between 1 and 2023 and $a$ divides $b$, then $a$ and $b$ are of different colors. What is the minimum number of colors Félix needs?