jmc

First, prime factorize $20^9$ as $2^{18}\cdot 5^9$. Denote $a_1$ as $2^{b_1}\cdot 5^{c_1}$, $a_2$ as $2^{b_2}\cdot 5^{c_2}$, and $a_3$ as $2^{b_3}\cdot 5^{c_3}$. In order for $a_1$ to divide $a_2$, and for $a_2$ to divide $a_3$, $b_1\le b_2\le b_3$, and $c_1\le c_2\le c_3$. We will consider each case separately. Note that the total amount of possibilities is $190^3$, as there are $(18+1)(9+1)=190$ choices for each factor. We notice that if we add $1$ to $b_2$ and $2$ to $b_3$, then we can reach the stronger inequality $0\le b_1<b_2+1<b_3+2\le 20$. Therefore, if we pick $3$ integers from $0$ to $20$, they will correspond to a unique solution, forming a 1-1 correspondence between the numbers $b_1$, $b_2+1$, and $b_3+2$. This is also equivalent to applying stars and bars on distributing the powers of 2 and 5 through differences. The amount of solutions to this inequality is $\left(\genfrac{}{}{0ex}{}{21}{3}\right)$. The case for $c_1$,$c_2$, and $c_3$ proceeds similarly for a result of $\left(\genfrac{}{}{0ex}{}{12}{3}\right)$. Therefore, the probability of choosing three such factors is$$\frac{\left(\genfrac{}{}{0ex}{}{21}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{12}{3}\right)}{190^3}.$$Simplification gives $\frac{77}{1805}$, and therefore the answer is $\boxed{77}$.