$a,b,c$ are three distinct natural numbers. Then how many ordered triplets $(a,b,c)$ will exist such that L.C.M (a,b,c) = 144.

Question

Let $a,b,c$ be three distinct natural numbers. Then how many ordered triplets $(a,b,c)$ will exist such that L.C.M (a,b,c) = 144.

Here's how I proceeded,

144=$(2^4)(3^2)$, so 144 has 15 factors(1 and 144 included). Let $F$ be the set of these factors. Then $a,b,c\in{F}$. I can't just select any 3 elements of $F$ as their LCM need not necessarily be 144. Is there any sort of further condition which may be applied on $a,b,c$ that will help me solve this problem.

Answer 1

Hint: the lcm of $2^i 3^j$, $2^k 3^\ell$ and $2^m 3^n$ is $2^{\max(i,k,m)} 3^{\max(j,\ell,n)}$.

Answer 2

At least one of them must have $2^4$ and at least one of them must have $3^2$.

Assign $3^2$ to any one and what choices do you have with others? Proceed similarly with $2^4$