$a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$

Question

How many pairs of ($a$,$b$) of positive integers are there such that $a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$?

What I tried

I know $6^6$ an be broken down into $(2)^6 (3)^6$

If $a$ is a factor of $b$,and if $a=1$,there will be $18$ groups.

Since $a$ cannot be higher than $n^3$ to be a factor of b,there are another $9$ groups.

So,total pairs must be $27$ right?

Answer 1

It suffices to consider powers of $2,3$ separately.

If $v_2(a)=r$ then $r≤v_2(b)≤6$. Of course, $v_2(a)\in \{0,1,2,3,4,5,6\}$. If $v_2(a)=0$ there are $7$ possibilities for $v_2(b)$. If $v_2(a)=1$ there are $6$ possibilities for $v_2(b)$, and so on. Thus, considering only powers of $2$, we get $$7+6+5+4+3+2+1=\frac {7\times 8}2=28$$ possible pairs.

The same calculation works for the powers of $3$.

As we can sort out powers of $2,3$ independently, we get $$28\times 28 =\fbox {784}$$