Argentina_2018

For $a\in \mathbb{N}$ denote $u(a)=\left[\frac{a}{2}\right]$, $v(a)=\left[\frac{a}{3}\right]$. If $n=p_1^{a_1}\cdots p_k^{a_k}$ is the prime factorization of $n\in \mathbb{N}$, it is straightforward that $D_2(n)=(u(a_1)+1)\cdots (u(a_k)+1)$, $D_3(n)=(v(a_1)+1)\cdots (v(a_k)+1)$.

Define $a_1=2\cdot 998$ and set $a_i=2v(a_{i-1})$ for $i\ge 2$. (Only the first several terms of the infinite sequence $(a_i)$ will be used.) With this definition we have $u(a_i)=998$ and $u(a_i)=v(a_{i-1})$ for $i\ge 2$.

Note also that $v(a_i)=\lfloor \frac{2}{3}v(a_{i-1})\rfloor <v(a_{i-1})$ if $v(a_{i-1})>0$. Thus the sequence $(v(a_i))$ decreases, so its terms are $0$ for sufficiently large $i$.

Take the first index $k$ such that $v(a_k)=0$ and define $n=p_1^{a_1}\cdots p_k^{a_k}$. Because $u(a_i)=v(a_{i-1})$ for $i\ge 2$,

$$D_2(n)=(998+1)(u(a_2)+1)\cdots (u(a_k)+1)=999(v(a_1)+1)\cdots (v(a_{k-1})+1).$$

In addition $D_3(n)=(v(a_1)+1)\cdots (v(a_{k-1})+1)(v(a_k)+1)$ equals $(v(a_1)+1)\cdots (v(a_{k-1})+1)$ since the last factor $v(a_k)+1$ is $1$ due to $v(a_k)=0$. Therefore $D_2(n)=999D_3(n)$.