Given LCM of three natural numbers, find the possibilities. LCM of three natural numbers =150. How many sets of three numbers are possible?
I know how to do this for two natural numbers.There is also a general formula for that. But for 3 numbers it is posing a problem. Is there a standard method for any n natural numbers?
Please help.
 A: I will assume that the numbers are distinct, and that the order doesn't matter.
If we let D be the set of positive divisors of 150, let S be the set of 3-element subsets of D, and 
let $A_i$ for $1\le i\le3$ be the set of 3-element subsets of D not containing a multiple of 2, 3, or 25, respectively,
then $|A_1^c\cap A_2^c\cap A_3^c|=|S|-|A_1|-|A_2|-|A_3|+|A_1\cap A_2|+|A_1\cap A_3|+|A_2\cap A_3|-|A_1\cap A_2\cap A_3|$
$\hspace{1.2 in}=\dbinom{12}{3}-\dbinom{6}{3}-\dbinom{6}{3}-\dbinom{8}{3}+\dbinom{3}{3}+\dbinom{4}{3}+\dbinom{4}{3}-0=133$.
A: Hint
We are looking for the LCM of $3$ numbers, so consider the prime factorization of $150$:
$$150 = 50*3  = 5^2*2*3$$
Therefore, we have numbers of the form: $$5^x2^y3^z = 150$$
Where $x$, $y$, and $z$ are integers.
We also know that: $
0 \le x \le 1$, $0 \le y \le 1$, $0 \le z \le 1$
A: $$150=2\cdot3\cdot5^2$$
The highest power of $2$ in at least one of the three number has to be $1$
This can be achieved in the following  ways: $\{1,0,0\};\{1,1,0\};\{1,1,1\}$
The highest power of $5$ in at least one of the three number has to be $2$
This can be achieved in the following  ways: 
$\{2,0,0\};\{2,1,0\};\{2,1,1\};\{2,2,0\};\{2,2,1\};\{2,2,2\}$
So, the number of possible distinct combinations $3\cdot3\cdot6$
