LCM of 3 numbers is given then what can't GCD be This question was asked to me by a junior and I was unable to solve it and in fact couldn't even start despite of having studied considerable number theory.

$\text{lcm}$ of three different numbers is $120$ then which of the following can't be it's HCF?

(A) $8$
(B) $12$
(C) $24$
(D) $35$
I have studied elementary number theory from David Burton but I am absolutely clueless about this particular problem.
I could think of $\gcd(a,b) \text{lcm}(a,b)=ab$ , but is this extendable  to more than two variables? and even if it is extendable it doesn't seem to be useful.
Thanks for any help!!
 A: Note that $\gcd(x_{1},x_{2},...,x_{n})\vert\ \text{lcm}(x_{1}, x_{2}, ..., x_{n})$ because $\gcd(x_{1},x_{2},...,x_{n})\vert\ x_{1}, x_{2}, ..., x_{n}$ and $x_{1}, x_{2},...,x_{n}\vert\ \text{lcm}(x_{1}, x_{2}, ..., x_{n})$. Thus, if a number does not divide the $\text{lcm}$, it cannot be the $\gcd$, so the answer must be $\boxed{(D)\ 35}$
A: Let lcm be $l$ and gcd (hcf) be $d$, $l = \dfrac{abc}{d}$, where $a, b, c$ are your three numbers.
Since $d|a, d|b, d|c$, $d$ must also divide divide $l$. Of the four answers, 35 is only one that does not divide 120.
A: The answer is D.  If $7$ is a divisor of each of the numbers, then it would also divide any common multiple.
A: We first observe that $120 = 2^4 \cdot 3^1 \cdot 5^2$. Additionally, we know that the gcd of any $n$ numbers must divide the lcm of those numbers.
Therefore, we see that choices $a$, $b$, and $c$ all divide the lcm. However, 35 does not divide the lcm and therefore it cannot be the gcd.
A: It's $(D)$ because $7$ is a factor of $35$ but it's not a factor of $120$
