Find the number of ordered triplets $x,y,z$ 
Find the number of ordered triplets $(x,y,z)$ of positive integers less than $13$ such that the product $x\cdot y\cdot z$ is divisible by $20$.

My work:
The possible values of $x\cdot y\cdot z$ are $20,40,60,80,100,120,140,160$. I began to make cases. I made cases for the product to be equal to $20$. I got $27$ cases.
Then for the product to be equal to $40$, I got $36$ cases.
I see that the number of cases are getting bigger. I don't know the best method of doing these types of questions.
Any help is greatly appreciated.
EDIT
Can we do it by the inclusion-exclusion principle$?$ Like counting all the triplets whose product does not divide both $5$ and $4$.
Here, $x,y,z$ have to be different as if it's not the case then the possible number of triplets exceed $216$.
 A: Here is a generating function approach. We consider the set of integers $\{1,2,3,\ldots,12\}$ and mark the occurrence of prime factors $2$ and $5$ according to their multiplicity with $z$ resp. $w$. We obtain
\begin{align*}
\begin{array}{cccccccccccc}
1&2&3&4&5&6&7&8&9&10&11&12\\
1&\color{blue}{z}&1&\color{blue}{z^2}&\color{blue}{w}&\color{blue}{z}&1&\color{blue}{z^3}&1&\color{blue}{zw}&1&\color{blue}{z^2}\\
\end{array}\tag{1}
\end{align*}
We obtain from (1) the generating function $A(z,w)$ as
\begin{align*}
\color{blue}{A(z,w)=5+\left(2z+2z^2+z^3\right)+(1+z)w}\tag{2}
\end{align*}
Since we want to count
\begin{align*}
|\{(x,y,z)\,:\,1\leq x,y,z\leq 12, 20|xyz\}|
\end{align*}
we calculate $A(z,w)^3$ which corresponds to the sum of products $x\cdot y\cdot z$ with $1\leq x,y,z\leq 12$ and count all terms which contain $z^2w$. These are the terms which correspond to a multiple of $20$.

Denoting with $[z^n]$ the coefficient of $z^n$ of a series we obtain
\begin{align*}
\color{blue}{\sum_{k\geq 2}}&\color{blue}{\sum_{l\geq 1}[z^k][w^l]A(z,w)^3}\\
&=\sum_{k\geq 2}\sum_{l\geq 1}[z^k][w^l]\left(5+\left(2z+2z^2+z^3\right)+(1+z)w\right)^3\tag{3.1}\\
&=\sum_{k\geq 2}[z^k]\left((1+z)^3+3\cdot 5(1+z)+3\cdot 5(1+z)^2\right.\\
&\qquad\quad+3\left(2z+2z^2+z^3\right)^2(1+z)\\
&\qquad\quad+3\left(2z+2z^2+z^3\right)\left(1+z\right)^2\\
&\qquad\quad+\left.6\cdot 5\left(2z+2z^2+z^3\right)(1+z)\right)\tag{3.2}\\
&=\sum_{k\geq 2}[z^k]\left(\left(3z^2+z^3\right)+3\cdot 5\cdot 0+3\cdot 5z^2\right.\\
&\qquad\quad+3(5)^2(2)\\
&\qquad\quad+3\left(\left(2z^2+z^3\right)+2\left(2z^2+2z^3+z^4\right)+\left(2z^3+2z^4+z^5\right)\right)\\
&\qquad\quad\left.+6\cdot 5\left(2z+2z^2+z^3\right)(1+z)\right)\tag{3.3}\\
&=4+0+15+150+3\left(3+2\cdot 5+5\right)+6\cdot 5\left(3+5\right)\tag{3.4}\\
&\,\,\color{blue}{=463}
\end{align*}
in accordance with other answers.

Comment:

*

*In (3.1) we consider a trinomial expansion
\begin{align*}
(a+b+c)^3&=a^3+b^3+c^3+3\left(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\right)\\
&\qquad+6abc
\end{align*}


*In (3.2) we select all terms which contains a factor $w$, i.e. which contains a factor $c$ in the expansion from $(a+b+c)^3$.


*In (3.3) we simplify terms by skipping constant and linear terms in $z$ and by counting some of the terms which contain a factor $z^k$ with $k\geq 2$.


*In (3.4) we simplify further until we finally derive the result.
A: I would condition on the number of x, y, z that are multiples of 5.
Let $F =\{4, 8, 12\}$; $E = \{2, 4, 6, 8, 12\}$; $O = \{1, 3, 7, 9, 11\}$ and $A=
\{1, 2, 3, 4, 6, 7, 8, 9, 11, 12\}$.
Then we can have
(a) Only 1 a multiple of 5. So $<5EE>$ (notation means a 5 and two elements of E), $<5FO>$, $<10EE>$, $<10EO>$.
$|<5EE>| = |E|\times 3 + {|E| \choose 2} \times 6$ (the two evens are equal, with 3 permutations of each or the two evens are distinct, with 6 permutations of each) = 75.
Similarly $|<5FO>| = |F|\times |O| \times 6$ = 90; $|<10EE>|$ = 75 and $|<10EO>|$ = 150. A total of 390.
(b) Exactly two are multiples of 5. So $<55F>$, $<510E>$, $<1010A>$, with $|<55F>| $ = 9; $|<510E>|$ = 30 and $|<1010A>|$ = 30. A total of 69.
(c) All three are multiples of 5. So $<101010>$ (only one), $<51010>$ (3 permutations). A total of 4.
Altogether 463.
