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!
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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.
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How many positive factors of $n$ are also multiples of $m$?
Unless $m$ is a factor of $n$, there aren't any.
But suppose $m\mid n$. Then a multiple of $m$ is $md$, and $md\mid n$ iff $n/(md)\in\Bbb N$ iff $d\mid(n/m)$. So, the answer to the question is "the number of positive factors of $n/m$".
On
How many positive factors of $x$ are also multiples of $y$?
None, unless $x$ is a multiple of $y$; so let's suppose that $\frac xy$ is an integer. In that case, $ky$ divides $x$ if and only if $y$ divides $\frac xy,$ so the number of positive factors of $x$ that are also multiples of $y$ is the same as the number of positive factors of $\frac yx,$ which I suppose you know how to compute from the prime factorization of $\frac yx.$
How many positive factors of $60$ are also multiples of $4$?
Since $\frac{60}4=15,$ that's the same as asking, how many positive integers are factors of $15$? The $4$ positive factors of $15,$ namely $1,3,5,15,$ correspond to the $4$ positive factors of $60$ which are multiples of $4,$ namely, $4,12,20,60.$
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: $2^2\cdot 3^0\cdot 5^0, 2^2\cdot 3^1\cdot 5^0, 2^2\cdot 3^0\cdot 5^1, 2^2\cdot 3^1\cdot 5^1$