How to find the number of trousers sold by a merchant on a specific day of the month? The problem is as follows:

The owner of a bazaar in Daegu obtained for the sale of pants of the
same price, $6804$ usd in September, $10800$ usd in October and $7938$
usd in December. If the price of each pants is greater than $14$ usd
and less than $21$ usd, how much did he earned on the first day of
January, assuming that he sold a number of pants equal to the product
of the digits of its price?.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&amp;\textrm{360 usd}\\
2.&amp;\textrm{252 usd}\\
3.&amp;\textrm{486 usd}\\
4.&amp;\textrm{144 usd}\\
\end{array}$
The official solution according to my precalculus workbook is as follows:
First, find the $\operatorname{gcd}$ of $6804$, $10800$ and $7938$.
Therefore:
$\operatorname{gcd}(6804, 10800, 7938)=54$
This can be obtained from the prime factorization of each number:
$6804=2^2\cdot 3^5\cdot 7$
$10800=2^4 \cdot 5^2 \cdot 3^3$
$938=3^4\cdot 7^2 \cdot 2$
Then $\operatorname{gcd}(6804, 10800, 7938)=2\cdot 3^3 = 54$
Therefore the pants price is $18$ usd.
Because the quantity sold on January 1st coincides with the product of the digits of the price then:
$\textrm{number of pants}=1\times 8=8$
Since the price per pant is $18$ usd. Then the quantity he earned is:
$8\times 18 =144\,\textrm{usd}$
Therefore the answer is option 5.
This is the part where it ends the official solution.
Now my problem with this answer is I don't know why did the author used the greatest common divisor? I mean at some point he does obtain $54$, but what is it used for? Can someone help me here?.
There is a very imporant aspect and missing step here and that is how did he got $18$ usd the price per pant?
I'm not very savvy with the correct usage of the greatest common divisor, but I believe the intented approach is that to get the price I must find a common divisor for the three. But why should this be the maximum?
The numbers which $54$ multiplies, with those being 126, 200 and 147 I assume that these must be the number of pants sold on the months of september, october and december. But from this logic I'm getting that the price of the pants is 54 usd.
This itself causes a contradiction as the boundaries set by the problem constrain the value of the price of the pants between:
$14&lt;p&lt;21$
If this were the case, then I believe that the price of the pants might be a sub multiple of 54?. But which?, 2?, 3?, 6?, 9?.
Because if any of those were the case the price could be, 15,16,18 and 20. This part confuses me the most.
So in brief the sort of answer which I'm aiming to get is a one which first and foremost explain why was it used the greatest common divisor then how was it obtained the price per pant and on which cases does the greatest common denominator is used? and how should it be avoided its confusion with the least common multiple?. In other words why using the least common multiple isn't used here. By the way I've read the wikipedia entry on the subject but it did not helped me much as it only covers two applications none of which seem to properly explain its usage in word problems.
Since what I'm lacking is the conceptual explanation it would really help me a step by step explanation for this problem. So can someone help me here?.
The price per pants must be a divisor of each of the amounts of money made in each of the months, as presumably he sold an integral number of pants. Thus, the price per pant must divide the greatest common common divisor of 6804 (money made in September), of 10800 (money made in October), and of 7938 (money made in December). Since the greatest common divisor is 54, the price per pant must be a divisor of 54.
(The defining property of the greatest common divisor of a set of numbers is that (i) it divides each of the numbers in the set; and (ii) any number that divides all numbers in the set divides the greatest common divisor. So the price per pant, which must divide each of the three amounts, must also divide $54$)
Now, what are the divisors of $54$? As $54=2\times 3^3$, the divisors are: $1$, $2$, $3$, $6$, $9$, $18$, $27$, and $54$. The price must be one of them.
We are told the price is “more than $14$ and less than $21$”. Well, the only possibility then is that the price per pant is $18$.
The way via the greatest common divisor is absolutely not necessary. Actually the revenues for the given three months aren't necessary at all to solve the problem. So, I add this short answer, even if it does not address the OP's question:
We know for the price $p=10a+b$ and $a\cdot b \in \{5,6,7,8,9\}$ since a price of $20$ would mean no pants sold.
Given the options, the revenue $p\cdot a\cdot b$ must be even. So only $16$ or $18$ are to be considered. Checking these, $p= 18$, hence, a revenue of $144$ is the only possible answer.