It is possible to solve this task "mathematically", by testing out certain cases. We leave the proof of this procedure as a practice for the reader and continue to explain a simpler solution.

We use a for loop to go through all possible values of the initial number of medallions $M$; for example, from $1$ to $1000$. For each such potential $M$ we check whether it is a possible solution. How? The youngest daughter would take $MdivN$ medallions, which means there are $M-MdivN$ medallions remaining. If that number equals the actual remainder $O$, we have found ourselves a solution. Additionally, we have to keep track of the smallest and biggest solution in two variables; we are going to output them in the end.

Please note: $div$ stands for integer division.