69th Belarusian Mathematical Olympiad

Answer: $n=3\cdot 673^{18}$; $n=3^{18}\cdot 673$.

Note that the equality $n=d_{19}\cdot d_{20}$ implies that $n$ has exactly $38$ divisors. Indeed, $$n=d_{19}\cdot d_{20}=d_{18}\cdot d_{21}=d_{17}\cdot d_{22}=\cdots =d_1\cdot d_{38}=d_{38}.$$ Hence either $n=p^{\alpha }$, where $p$ is prime and then $\alpha +1=38$, so $n=p^{37}$, or $n=p^{\alpha }\cdot q^{\beta }$ ($p$ and $q$ are different primes) and then $38=(\alpha +1)(\beta +1)$, i.e. $(\alpha ;\beta )=(1;18)$ or $(\alpha ;\beta )=(18;1)$, so $n=p\cdot q^{18}$ or $n=p^{18}\cdot q$, respectively.

Recall that $n$ is divisible by $2019$ while $2019=3\cdot 673$, where $3$ and $673$ are primes. Therefore, $n=3\cdot 673^{18}$ or $n=673\cdot 3^{18}$. It remains to verify that in both of these cases $n=d_{19}\cdot d_{20}$.

1. Let $n=3\cdot 673^{18}$. Arrange the divisors in an ascending order: $$1<3<673<673\cdot 3<673^2<673^2\cdot 3<\cdots <673^9<673^9\cdot 3<\dots$$ It is easy to see that $d_{19}=673^9$, $d_{20}=673^9\cdot 3$ and $n=d_{19}\cdot d_{20}$.

2. Let $n=3^{18}\cdot 673$, then $$\begin{gathered} 1<3<3^2<3^3<3^4<3^5<673<3^6<3\cdot 673<3^7<3^2\cdot 673<3^8<3^3\cdot 673< \\ <3^9<3^4\cdot 673<3^{10}<3^5\cdot 673<3^{11}<3^6\cdot 673<3^{12}<\dots \end{gathered}$$ It is easy to see that $d_{19}=3^6\cdot 673$, $d_{20}=3^{12}$ and $n=d_{19}\cdot d_{20}$.