AUT_ABooklet_2020

Numbers with exactly 2019 positive divisors are either of the form $p^{2018}$ or $p^{672}\cdot q^2$ for distinct prime numbers $p$ and $q$. The number $128^{97}$ can be written as $$128^{97}=(2^7{)}^{97}=2^{679}.$$ As $p$ is at least $2$, the number $p^{2018}$ is greater than $2^{679}$ and therefore, the case $n=p^{2018}$ is impossible. Thus we have $n=p^{672}\cdot q^2$ with $p^{672}\cdot q^2<2^{679}$. Hence $p=2$ and as $q^2<2^7=128$, $q$ is one of the primes $3$, $5$, $7$ or $11$.

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 $n=2^{672}\cdot 11^2$.