Exercise 13. For any integer $n \geqslant 0$, we name $s(n)$ the sum of the digits of $n$. Determine all integers $n \geqslant 0$ such that $n \leqslant 2 s(n)$.
Solution to Exercise 13 Let $n$ be a natural number and let $a_{d} a_{d-1} \ldots a_{1} a_{0}$ be its decimal representation, meaning that $a_{0}$ is the units digit, $a_{1}$ is the tens digit, etc...

First, if $n$ has only one digit, then $n=a_{0} \leqslant 2 a_{0}$, so $n$ is a solution to the exercise, and the numbers $0,1,2, \ldots, 9$ are solutions.

We now assume that $n$ has at least two digits.

Then $n=a_{d} \cdot 10^{d}+a_{d-1} \cdot 10^{d-1}+\ldots a_{1} \cdot 10^{1}+a_{0} \cdot 10^{0}$ and $s(n)=a_{d}+\ldots+a_{1}+a_{0}$. If $d \geqslant 2$, then for all $1 \leqslant i \leqslant d-1, a_{i} \cdot 10^{i} \geqslant a_{i} \cdot 2$ and since $a_{d}$ is non-zero, $a_{d} \cdot 10^{d}=2 a_{d}+\left(10^{d}-2\right) \cdot a_{d} \geqslant 2 a_{d}+10^{d}-2>2 a_{d}+a_{0}$ $\operatorname{since} a_{0}2 a_{d}+a_{0}+2 a_{d-1}+\ldots+2 a_{1}+a_{0}=2 s(n)
$$

Therefore, if $n$ has three digits or more, $n$ is not a solution.

We assume that $n$ has two digits. If $a_{1} \geqslant 2$, then

$$
n=a_{1} \cdot 10+a_{0}=2 a_{1}+a_{0}+8 \cdot a_{1} \geqslant 2 a_{1}+a_{0}+8 \cdot 2>2 a_{1}+a_{0}+a_{0}=2 s(n)
$$

so $n$ is not a solution to the problem. We deduce that $a_{1}=1$. We can then easily test all the numbers between 10 and 19 to see which ones satisfy the problem. Or, we can note that the inequality $n \leqslant 2 s(n)$ can be rewritten as $10 \cdot 1+a_{0} \leqslant 2\left(1+a_{0}\right)$, which simplifies to $a_{0} \geqslant 8$; therefore, the only two-digit numbers that satisfy the statement are 18 and 19. Conversely, we have $18 \leqslant 2(1+8)$ and $19 \leqslant 2(1+9)$.

The solutions sought are therefore $\{0,1,2,3,4,5,6,7,8,9,18,19\}$

Comment from the graders: The problem is well understood, students quickly have an intuition of the result and the reasoning to be done, but it is more difficult to set it up rigorously. The upper bound of $s(n)$ was a good idea. It is important to avoid parachuting results, even if they are correct, when they are not immediate: in this case, proving that for all $k \geqslant 3$, we have $18 k<10^{k-1}$ was one of the main difficulties of the exercise, so a proof of this result was expected. Finally, many students forget that 0 is a solution.