Let $a$ be a factor of $120$ then what are the total number of positive integral solutions to $xyz=a$ including 120. The answer is $320$ . After wasting almost $15$ mins in getting the factors of each of its factors I got the answer. But the method is lengthy like writing $24=2.2.6=4.6.1=4.3.2=12.2=8.3.1=24.1.1$ . So I was interested in a generalized formula for total solutions to $xyz=n$ with or without repetitions.
2026-04-18 23:34:18.1776555258
Total number of integral solutions to the factors of a given numbet
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Equivalently we can look at solutions of $xyzw=120$ (where $w=\frac{120}{xyz}$ is uniquely determined by $x,y,z$). Also $120=2^3\cdot 3\cdot 5$.
There are four choices for which variable gets the factor of $3$; and four choices for which variable gets the factor of $5$. And using 'stars and bars' there are ${6\choose 3}=20$ choices for how the factors of $2$ go to the variables.
Putting this together gives your answer of $320$.