ChallengeMediumHard Number Theory ShareHard Number Theory · Medium problemFind the number of trailing zeros (in base 101010) of the product ∏k=120k!=1!⋅2!⋯20!\displaystyle\prod_{k=1}^{20} k! = 1!\cdot 2!\cdots 20!k=1∏20k!=1!⋅2!⋯20!.Check Show answer Walk through it step by step Practice more like this Review this concept