Exercise 8. How many integers $n$ are there such that $\frac{n}{3}$ and $3 n$ are both integers between 1 and 1000 (inclusive)?

Only a numerical answer is expected here.
Solution to Exercise 8 If $\frac{n}{3}$ is an integer, let's call this integer $k$. Then we must have $1 \leqslant k \leqslant 1000$ and $1 \leqslant 9 k \leqslant 1000$.

Conversely, if we have an integer $k$ such that $1 \leqslant 9 k \leqslant 1000$, then setting $n=3 k$ produces an $n$ that works. Thus, there are as many such $n$ as there are integers that satisfy $1 \leqslant k$ and $9 k \leqslant 1000$.

There are $\left\lfloor\frac{1000}{9}\right\rfloor=111$.

Alternative Solution $n^{\circ} 1$ For $\frac{n}{3}$ to be an integer, $n$ must be divisible by 3. Furthermore, $3 n \leqslant$ 1000, so $1 \leqslant n \leqslant 333$.

Conversely, all multiples of 3 between 1 and 333 work, and there are 111 of them.

Grader's Comment: The exercise was very well done. Unfortunately, some students who did not have the correct result did not explain their reasoning and therefore did not receive any points.