Determine all natural numbers n that are less than $128^{97}$ and have exactly 2019 divisors.

(Richard Henner)

Answer. There are 4 solutions: $n=2^{672} \cdot 3^{2}$ or $n=2^{672} \cdot 5^{2}$ or $n=2^{672} \cdot 7^{2}$ or $n=2^{672} \cdot 11^{2}$.
Let $p$ and $q$ be prime numbers. Numbers that have exactly 2019 divisors are either of the form $p^{2018}$ or $p^{672} \cdot q^{2}$. The number $128^{97}$ can be represented as

$$
128^{97}=\left(2^{7}\right)^{97}=2^{679} \text {. }
$$

The number $p^{2018}$ is of course greater than $2^{679}$, because $p$ is at least 2. Therefore, it must hold that $n=p^{672} \cdot q^{2}<2^{679}$. Hence, $p=2$ and $q^{2}<2^{7}=128$ must be true. Therefore, the only primes that can be $q$ are 3, 5, 7, or 11.

Thus, there are 4 solutions: $n=2^{672} \cdot 3^{2}$ or $n=2^{672} \cdot 5^{2}$ or $n=2^{672} \cdot 7^{2}$ or $n=2^{672} \cdot 11^{2}$.

(Richard Henner)