2021 Fall AMC 10B Problem 6

Before starting, note that if we can represent the prime factorization of an integer $z$ as $$z=p_1^{e_1} p_2^{e_2} \cdots,$$ then there are $(e_1+1)(e_2+1) \cdots$ distinct positive factors. If the number in question has $2021$ factors, by the previous logic, $$2021=(e_1+1)(e_2+1) \cdots,$$ and as the prime factorization of $2021 = 43\cdot 47,$ then our number must be $p_1^{46}p_2^{42}$ or $p^{2020}.$ The smallest number we can make in either of these is making $p_1 = 2,p_2=3$ in the first configuration, yielding $$2^{46}3^{42} = 16\cdot 6^{42} .$$ Therefore, $$m=16$$$$k=42,$$ so $$m+k = 42+16 = 58.$$ Thus, the answer is B .