Let $n \geqslant 1$ be an integer. Morgane initially has $n$ piles, each containing one coin. She then allows herself operations of the following form: she chooses two piles, takes as many coins from the first pile as from the second, and forms a new pile with the coins she has taken.
Determine, as a function of $n$, the smallest number of non-empty piles that Morgane can obtain from such operations.
If $n$ is a power of 2, Morgane can manage to get only one pile by proceeding as follows: she merges the $n$ piles of size 1 into $n / 2$ piles of size 2, then merges these $n / 2$ piles into $n / 4$ piles of size 4, and so on.
Conversely, when Morgane extracts $k$ pieces from two piles originally containing $a$ and $b$ pieces to form three piles (possibly empty) of $a-k, b-k$, and $2k$ pieces, we observe that $\operatorname{PGCD}(a-k, b-k, 2k)$ divides $2a$ and $2b$, and thus divides $2 \operatorname{PGCD}(a, b)$. Consequently, an immediate induction shows that, if Morgane has $k$ piles of sizes $t_{1}, t_{2}, \ldots, t_{k}$, the number $\operatorname{PGCD}\left(t_{1}, t_{2}, \ldots, t_{k}\right)$ is a power of 2. In particular, if Morgane manages to have only one pile, which will contain $n$ pieces, it follows that $n$ is a power of 2.
Finally, here is how, for any value of $n$, Morgane can proceed to end up with only two piles. She writes $n$ as a sum of increasing powers of 2, that is, numbers $2^{a_{1}}<2^{a_{2}}<\cdots<2^{a_{k}}$ such that $n=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{k}}$. For each $i \leqslant k$, and proceeding as in the case where $n$ is a power of 2, Morgane groups $2^{a_{i}}$ of her pieces into a single pile, thus obtaining $k$ piles.
Morgane then distinguishes her pile of size $2^{a_{1}}$, which she calls the target, and her pile of size $2^{a_{k}}$, which she calls the reserve. The $k-2$ other piles are seen as normal piles. Then Morgane sets out to double the size of her target pile while ensuring it will never be larger than a normal pile, proceeding as follows:
$\triangleright$ if the target pile is the same size as a normal pile, Morgane merges these two piles into a new target pile; a normal pile disappears during this operation;
$\triangleright$ otherwise, the target pile is of size $2^{a}$ strictly smaller than each normal pile, and Morgane then extracts $2^{a}$ pieces from the reserve to integrate them into the target pile, which is now of size $2^{a+1}$.
After $\ell$ such steps, the target pile is now of size $2^{a_{1}+\ell}$, and it has absorbed the piles of sizes $2^{a_{i}}<2^{a_{1}+\ell}$, as well as at most $2^{a_{k-1}}$ pieces from the reserve, which is therefore still not empty. Thus, after $a_{k-1}+1-a_{1}$ steps, the target pile is now of size $2^{a_{k-1}+1}$, and it has absorbed all the normal piles, which means that one is left with only two piles.
In conclusion, the minimum number of piles Morgane can end up with is 1 if $n$ is a power of 2, and it is 2 otherwise.

Comment from the graders The exercise was of progressive difficulty. Like many combinatorial exercises, it rewarded students who tested small cases. Indeed, by testing small cases, such as 2 and 4, one sees that if $n$ is a power of 2, then one can form a single pile. By continuing with small cases such as 3, 5, and 6, one sees that if $n$ is a sum of $k$ powers of 2, then one can make $k$ piles. But $k$ is not necessarily the minimum: for 7, one can obtain a pile of 4, one of 2, and one of 1; then one can obtain a pile of 3, and two of 2, thus a pile of 4 and one of 3. Thus, one can do better than the number of 1s in the binary representation of $n$.
By testing numbers from 8 to 16, one sees that one can obtain 2 piles for numbers that are not powers of 2, but not better, and that the good strategy is to make a large pile of a power of 2, then to slightly empty it to form 2 piles. After that, it only remains to formalize this argument to show that one can obtain two piles.
The most difficult part of the exercise remained: why, if $n$ is not a power of 2, one cannot obtain a single pile. Unfortunately, some misread the statement and thought one had to remove as many pieces from the first pile as there were in the second: this is not stated in the statement. Some tried to proceed by induction to show that several steps before the end, there was a pile of size $n / 2^{k}$ (or even $2^{k}$ such piles). This is not true for $k=3$: indeed, one can imagine having a pile of $5 n / 8$ and one of $3 n / 8$, taking $n / 8$ pieces from each to have a pile of $n / 2$ and two of $n / 4$, and in two merges obtaining a single pile.
The argument was more subtle: the simplest was to write $n=2^{k} \times y$ with $y$ odd, and to show by induction that $k$ steps before the end, each pile had a size divisible by $y$. Alternatively, to show that the piles had a size of the form $k n / 2^{j}$ with $k$ odd, and thus could never be of size 1.