Problem 2. How many divisors of the number $2021^{2021}$ have a cube root that is a natural number?
Answer: 454276.

Solution. Since $2021=43 \cdot 47$, all divisors of the number $2021^{2021}$ have the form $43^{\alpha} \cdot 47^{\beta}$, where $\alpha, \beta \in[0 ; 2021]$. In this case, the exact cubes are numbers of the form $43^{3 n} \cdot 47^{3 k}$, where $3 n, 3 k \in[0 ; 2021]$, that is, $n, k \in[0 ; 673]$. There are $674^{2}=454276$ such numbers.