number of interesting divisors of $9!$ 
We call natural number $a$, "interesting" if natural numbers $x,y$
(not necessary distinct) exist so that $a=x^2y^3$. how many
interesting divisors the number $9!$ has?

To avoid calculations errors question brings this hint: "sum of the digits of the answer is divisible by $10$."
Here is my work:
for $x=y=1$ we have $a=1$ so it is one of the number we are looking for. also we know that  $9!=2^7\times3^4\times5\times7$, from this, we can conclude that $x$ and $y$ can't be $5$ or $7$ (because their exponent is $1$) so we should working on $2^7\times3^4$ and (also should considering $1$).
I don't know how to continue from here. I tried counting all the answers but my result is $16$ which is wrong. (maybe I missed some cases and answer should be $19$ according to the hint)
EDIT:
I got these answers:
$$x=2 , y=1,3$$
$$x=4 , y=1,3$$
$$x=8, y=1,3$$
$$x=3, y=1,2,4$$
$$x=9,y=1,2,4$$
$$x=1, y=1,2,3,4$$
 A: Suppose that $p^r$ is an exact prime power divisor of $a^2b^3$ then if the exact prime power divisors of $a$ and $b$ are $p^s$ and $p^t$ respectively, then $$r=2s+3t$$
Also note that if $r=2s+3t$ then $p^r=(p^s)^2(p^t)^3$
So the exponent for every prime has to be of the form $2s+3t$ and any non-negative exponent of this form will do.
That should help your counting.
A: This question is trickier than it appears. There are $28$ interesting divisors. Rather than work everything out in detail, I will give sufficient hints so that you can enjoy the hunt.
$x$ can be selected from $1,2,3,4,6,8,9,12,18,24,36$. For each $x$, there are one or more possible values of $y$, including $1$. I leave it to you to discover each possibility. When you count up all the possibilities, you will find $32$ combinations of $(x,y)$ that are permitted.
Here's the tricky part: three of the divisors $(64,576,1728)$ can be arrived at in more than one way; when you remove duplicates, you are left with the $28$ interesting divisors. If you have trouble finding them after these hints, ask in a comment for the list and I will put it in the answer by an edit.
