By Legendre $$v_2(50!)=\left \lfloor \frac{50}{2^1} \right \rfloor + \left \lfloor \frac{50}{2^2} \right \rfloor + \left \lfloor \frac{50}{2^3} \right \rfloor + \left \lfloor \frac{50}{2^4} \right \rfloor + \left \lfloor \frac{50}{2^5} \right \rfloor=25+12+6+3+1=47$$ Thus, the answer is $n=47$

How is this on the olympiad?

parmenides51 wrote: What is the maximum integer $n$ such that $\frac{50!}{2^n}$ is an integer? how is this an old high school olympiad (2020)?

all those problems are posted in order to complete the contest collections, even if they are recent ones, I cannot post at once 30 problems from a contest in HSO as it would be considered as spamming, therefore I choose which to problem to post in each forum and why after admins add those post collections in contest collections, everyone will find them at once