The number of ordered pairs $(p, q)$ of positive integers, whose $\operatorname{lcm}$ is $8100$

Question

The number of ordered pairs $(p, q)$ of positive integers, whose $\operatorname{lcm}$ is $8100$, is $K$. Then find the number of ways of expressing $K$ as a product of two distinct positive integers.

Now I have taken the $\text{lcm} (p,q)=8100$ which can be expressed as $2^2 \times 5^2 \times 3^4$.

After this, I have taken as $p$ to be $2^a 5^b3^c$ and $q$ to be $2^x5^y3^z$ where $a,b,c,x,y,z$ here represent non negative integers. From there I have taken $\max\{a, x\}= 2$, $\max\{b, y\} = 2$, $\max\{c, z\} = 4$. However from here, I am unable to proceed any further.

Answer 1

The number of ordered pairs of nonnegative integers $(a,x)$ with $\max(a,x)=2$ is $5$. This is because if $a=2$ there are $3$ possible choices for $x$ and if $x=2$ there are $3$ possible choices for $a$; this covers all cases but counts the pair $(2,2)$ twice, so there are $2\times 3-1$ pairs. From here you should be able to complete the solution.

Answer 2

I'm using "ordered" (as you clarify in the comments) to mean that the order of $(p,q)$ is important and makes a distinct pair, unless $p=q$.

Let's work up to this by solving for $4=2^2$ then $36=2^2\cdot 3^2$.

So the number of $(p,q)$ for $\text{lcm}(p,q)=4$ relies on having one of $p,q=4$, and the other can be $\in \{1,2,4\}$ - that is, $2+1=3$ options from the exponent going from $0$ to $2$. This gives us $3\times 2 = 6$ solutions (the ordering giving the $\times 2$), but then we remove the double counted $(4,4)$ giving $5$.

Now for $36$: We can solve independently as above for the powers of $2$ and $3$ but the only duplicate will be $(36,36)$ since no other repeated number will give lcm of $36$. So we have $6\times6-1 = 35$ solutions. (If we wanted "unordered" $(p,q)$, meaning $(18,4)\equiv (4,18)$, we would divide by $2$ rather than subtract $1$.)

Now it's clear that the number of solutions for $8100=2^2\cdot3^4\cdot5^2$ is $6\times 10\times 6 - 1 = 359$ options for $(p,q)$.

And, for the follow-on part of the question, $K=359$ is a prime number.