Question: I recently came upon the question that follows. 
“How many positive factors of 60 are also multiples of 4?”
Since 60 is relatively small I used a brute force method.
But I was wondering if there was a way you could do this using the prime factorization of 60.

Thanks!

Answer: 
Yes, indeed you can. Suppose that $a\mid b$ and you ask how many factors of $b$ are also multiples of $a$. Suppose $a=p\_1^{a\_1}p\_2^{a\_2}\cdots p\_k^{a\_k}$ and that $b=p\_1^{b\_1}p\_2^{b\_2}\cdots p\_k^{b\_k}p\_{k+1}^{b\_{k+1}}\cdots p\_n^{b\_n}$ (*where the primes are potentially written out of order to make it so all primes appearing in $a$ occur before those which don't appear in $a$*).

There will be $(b\_1-a\_1+1)(b\_2-a\_2+1)\cdots(b\_k-a\_k+1)(b\_{k+1}+1)\cdots (b\_n+1)$ number of such factors of $b$ which are also multiples of $a$.

This can be seen directly by [multiplication principle](https://en.wikipedia.org/wiki/Rule_of_product).

For the specific case of $4$ and $60$, note $4=2^2$ and $60=2^2\cdot 3^1\cdot 5^1$ so there are $(2-2+1)(1+1)(1+1)=1\cdot 2\cdot 2 = 4$ such factors of $60$ which are also multiples of $4$. Namely $4,12,20,60$, or rewritten: $2^2\cdot 3^0\cdot 5^0, 2^2\cdot 3^1\cdot 5^0, 2^2\cdot 3^0\cdot 5^1, 2^2\cdot 3^1\cdot 5^1$
Sure there is. It's more instructive to look at a bigger example. How many factors of $720$ are multiples of $4?$ Since $720=2^4\cdot3^2\cdot5,$ any factor of $720$ is of the form $2^a3^b5^c,$ where $0\le a\le4,\ 0\le b\le2,\ 0\le c\le1.$ However, for it to be a multiple 0f $4,$ we need $2,\le a$, so we have $3$ choices for $a$, $3$ for $b,$ and $2$ for $c,$ or $3\cdot3\cdot2=18$ in all.