# 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!!

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 $$\boxed{(D)\ 35}$$

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.

The answer is D. If $$7$$ is a divisor of each of the numbers, then it would also divide any common multiple.

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.

It's $$(D)$$ because $$7$$ is a factor of $$35$$ but it's not a factor of $$120$$