Let $S$ be the set of all positive integers $n$ s.t. $n^2$ is a multiple of both $24$ and $108$. Which of the following integers are divisors of every integer $n$ in $S$?
Attempt:
$n^2=24k \wedge n^2=108j; k,j \in \mathbb{Z} $
$n^2=3*2^3k \wedge n^2=3^2*2^2j$
$n=2*\sqrt{6k} \wedge n=6j$
I don't know where to go from here.
The answer choices are
- A:12
- B:24
- C:36
- D:72
$n^2=3*2^3k \wedge n^2=3^3*2^2j$
For $n^2$ to be a multiple of both $3*2^3$ and $3^3*2^2$, it has to be a multiple of $3^3*2^3$(Just take the highest prime powers). The minimum $n$ for this to happen is $n=3^2*2^2$. Every integer in $S$ is a multiple of $3^2*2^2$.