The problem is as follows:
The number of panadol pills at a pharmacy is a positive whole number that it has two prime divisors and 45 positive divisors. The number of tylenol pills at the same pharmacy is also a positive whole number which has 66 positive divisors. Assuming that both numbers which represents panadol and tylenol pills have the same prime divisors. How many compount divisors will it have the smallest value of the product of both numbers?
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{210}\\ 2.&\textrm{214}\\ 3.&\textrm{187}\\ 4.&\textrm{207}\\ \end{array}$
The official solution according to my book is to assume the following:
It states:
$\textrm{A: The number of panadol pills}$ $\textrm{B: the number of tylenol pills}$
$A=p^a q^b$ $B=p^c q^d$
According to this in order to get the minimum product of pills then $A\times B$ must be minimum.
For this to happen:
$\textrm{p<q and a>b and c>d}$
(Comment: This part it doesn't seem much sense)
Then it states: Using the notation of $\textrm{QD = quantity of divisors}$
$\operatorname{QD}(A)=(a+1)(b+1)=45$
$\operatorname{QD}(B)=(c+1)(d+1)=66$
From this it concludes that:
$a=9$
$b=4$
Assuming that the combination can only be: $9\times 5=45$
It doesn't say why it discards other choices.
and for the other number it assumes:
$22\times 3 =66$
Therefore
$c=21$
$d=2$
Again it doesn't say why it does have to be those numbers and in that order.
Then it declares:
$A\times B = p^8 \cdot q^4 \times p^{21}\cdot q^2= p^{29} \cdot q^6$
Now it comes the most weird conclusion:
$\operatorname{QD}(A\times B)= 19\times 10=190$
Hence:
It uses the formula:
$\operatorname{QD}=\textrm{prime divisors}+1+\textrm{compount divisors}$
I believe this arises from the fact that the number of divisors is equal to the number of prime divisors plus one plus the compound divisors.
Then:
$190=2+1+\textrm{compount divisors}$
Therefore it concludes:
$\textrm{compound divisors}=187$
Hence the answer is the third option.
But this doesn't make any sense. I mean at the point I reached this part I got lost.
As it can be seen I don't feel happy with this method of solution because it lacks of the proper explanation on why it must be those peculiar divisors it could be any other combination as well. This confuses me a lot.
As I mentioned it isn't very clear why that must be the quantity of divisors and that the smallest number of pills.
Does it exist a way to solve this using those concepts but better explained?
I'm lost at this ans since I feel there are many missing steps it would help me a lot if someone could help me here with an answer which would be easier to understand and the most detailed as possible.
First you need to be familiar with the fact that the divisors of $p^aq^b$ are all the numbers of the form $p^xq^y$ with $0\le x\le a, 0\le y\le b$. There are $(a+1)(b+1)$ of these.
So we know that $$(a+1)(b+1)=45$$ $$(c+1)(d+1)=66.$$
Now $45=1\times 45=3\times 15=5\times 9=9\times 5=15\times 3=45\times 1$. The possible solutions of the first equation are $$(a,b)\in \{(2,14),(4,8),(8,4),(14,2)\}.$$
By the same method $$(c,d)\in \{(1,32),(2,21),(5,10),(10,5),(21,2),(33,1)\}.$$
There are of course lots of ways of combining these. You should have a go at trying this out and for each possibility working out the number of divisors of the product of the two numbers, $(a+c+1)(b+d+1)$.
You can see that one possibility is $$(a,b)=(8,4)\text { and } (c,d)=(10,5)$$ This gives a number of divisors $19\times10=190$. Now, three of the divisors are $ 1,p,q$ and these are not composite and so we have the 'book' solution of $187$ composite factors.
Note
I don't think you should expect to be able to 'see' that $(a,b)=(8,4)\text{ and } (c,d)=(10,5)$ is obviously best. However if you try out some of the other possibilities you will get a feel for whether any other combination is likely to be better.
For example
$(a,b)=(2,14)\text { and } (c,d)=(2,21)$ gives $5\times 36=180$, a better answer! However the wording of the question concerning what is 'smallest' might mean that the product itself is smallest (although we don't know the primes). Do you have an image of the actual question?