GRE test prep question [LCM and divisors]

Question

Let $S$ be the set of all positive integers $n$ such that $n^2$ is a multiple of both $24$ and $108$. Which of the following integers are divisors of every integer $n$ in $S$ ?

Indicate all such integers:

$A:12$

$B:24$

$C:36$

$D:72$

The answers are $A$ and $C$

First I took the lcm of $24$ and $108$ which is $2^3\times3^3$ but then it says that "the prime factorization of a square number must contain only even exponents. Thus, the least multiple of $(2^3)(3^3)$ that is a square is $(2^4)(3^4)$"

Can somebody explain why that is true?

What if the lcm was $2^3\times3^4$ ? Would I just make it $2^4\times3^4$ ?

Help!

Answer 1

Let $n=p_1^{a_1}\cdots p_r^{a_r}$ where the $p_i$ are primes, so $n^2=p_1^{2a_1}\cdots p_r^{2a_r}$. As you observed, $n^2$ must be a multiple of $LCM(24, 108)=2^{3} 3^{3}$, so $2a_1\ge 3$ and $2a_2\ge 3$ with $p_1=2$ and $p_2=3$. Therefore $a_1\ge 2$ and $a_2\ge 2$, so n is a multiple of $2^{2} 3^{2}=36$. Thus S consists of all positive multiples of 36, so the integers which divide every integer in S are simply the divisors of 36.

Answer 2

We know that solution set is multiple of LCM . But here n^2 must be multiple of LCM . For that to happen it can be the number must be a perfect square . That can happen only if the power is even . Or else we will end up with square root in n .

In your case $ 2^3 * 3^4 $ as $n^2$. n will have $\sqrt{2}$ which makes n a non integer . so the numbers must be 216*6 , 216*6^3 ,...