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 ~M \mathbin{div} N~ medallions, which means there are ~M - M \mathbin{div} N~ 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.