<s>[INST] Formulas for solving problems such as “how many positive factors of x are also multiples of y?” 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!
 [/INST] 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.
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: $2^2\cdot 3^0\cdot 5^0, 2^2\cdot 3^1\cdot 5^0, 2^2\cdot 3^0\cdot 5^1, 2^2\cdot 3^1\cdot 5^1$
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