Fall 2021 AMC 10 B

The number of positive integer divisors of the positive integer whose prime factorization is $p_1^{e_1}{p}_2^{e_2}\cdots p_n^{e_n}$ equals $(e_1+1)(e_2+1)\cdots (e_n+1)$. Because $$2021=2025-4=45^2-2^2=(45+2)(45-2)=47\cdot 43,$$ a number having $2021$ divisors must be of the form $p^{2020}$ or $p^{46}\cdot q^{42}$, where $p$ and $q$ are distinct primes. This is minimized by taking $p=2$ in the first case, and $p=2$ and $q=3$ in the second case. Because $$2^{46}\cdot 3^{42}<2^{46}\cdot 4^{42}=2^{46+2\cdot 42}=2^{130}<2^{2020},$$ the least such positive integer is $$2^{46}\cdot 3^{42}=2^4\cdot 2^{42}\cdot 3^{42}=2^4\cdot 6^{42}=16\cdot 6^{42}.$$ Therefore $m+k=16+42=58$.