Word problem relating the divisors of a product 
Let $A$ be an arbitary finite set of positive integers. Anna and Bella play the following game. First Anna chooses a number $a \in A$, but does not tell it to Bella. Bella then chooses an integer $b$ (which does not have to be an element of $A$). Then Anna tells Bella the number of divisors of $ab$. Prove that Bella can choose $b$ in such a way that she may uniquely identify the number chosen by Anna.

This seemed to be a nice problem. I tried to get an intuition for it by trying it out with some specific values, but didn’t quite get it.
If we let $a=8$ and suppose that $d(ab) = 4.$ It seems that from here Bella could let $b = 1$ and then find integers that have $4$ divisors, but this woudn’t work at all. What would be the way to go about this?
 A: A good hint for solving this problem is the one that user376343 gave you: 
$$ \text{Study prime divisors of the element in } A \text{ and their powers and try to think at a suitable } B$$
Another hint, that probably you already know, is that if you have an integer number 
$$ c = p_1^{n_1} p_2^{n_2}\dots p_l^{n_l}$$
then 
$$ div(c) = (n_1+1)(n_2+1) \dots (n_l+2) $$
Now I'll write the answer. (spoiler alert)  
Express all the numbers in $A$ as products of prime in the form $p_1^{n_1}\dots p_l^{n_l}$.  
Now make a list of all the primes $\{q_i\}$ that appear at least in one of the numbers and make a list $\{m_i\}$ where $m_i$ is the greatest power of $q_i$ that divide an element in $A$, form this list take $M$ the maximum number that appears. 
Choose the lowest prime $q_1$ and associate it the number $q_1^{M^{l}-1}$.
Choose the next prime $q_2$ and associate it the number $q_2^{M^{2l}-1}$.
Continue like this and associate to the prime $q_i$ the number $q_i^{M^{2^{i-1}l}-1}$.
Now you can take $b$ equal to the product of all this numbers $b=\prod_i q_i^{M^{2^{i-1}l}-1}$.
Now you calculate $div(ab)$ where $a=q_1^{n_1} \dots q_l^{n_l}$ (you express $a$ as the product of all the $q_i$ of the list, if necessary some of the $n_i$ are equal to $0$) and $b$ is the number you just found.
Using the formula in the hint you know that $$div(ab) = (M^{l}+n_1)(M^{2l}+n_2) \dots (M^{2^{l-1}l}+n_l) $$
At this point you can watch the rapresentation in base $M^l$ of the number and you will read $n_i$ at the "digit" of place $1+2+\dots+2^{l-1}-2^{i-1} $.
When you solve the product, every addend will be in a different "digit" in base $M^l$, therefore you can reconstruct every $n_i$ and reconstruct the $a$ chosen by Anna that will be $p_1^{n_1} \dots p_l^{n_l}$.  

Explaining why this happens is a little tricky (hoping that I'm not wrong).  
When you solve the product $(M^l+n_1)(M^{2l}+n_2)\dots(M^{2^{l-1}l}+n_l) $ you  have the sum of different addend in the following shape $ n_{i_1} \dots n_{i_k} M^{2^{j_1}l} \dots M^{2^{j_{k'}l}} $ .
Since $n_{i_1} \dots n_{i_k} \le M^{l-1} < M^l $ it has no effect on the "digit" that this addend modifies, so the only intresting part is $ M^{2^{j_1}l} \dots M^{2^{j_{k'}}l}$ ; but since this is sum of different powers of 2 this will be uniquely defined and will never happen that two different addend has the same exponent of $M$.
