Proving there are $\left\lfloor\frac{n}{\text{lcm}(a,b)}\right\rfloor$ numbers divisible by a and b in [1,n]

Question

I am trying to prove the following (Please let me know if it isn't actually true):

Consider the range $[1,n]$ where $n \in \mathbb{Z}^+$. Consider $a,b \in \mathbb{Z}^+$.Prove that, in this range, there are $\left\lfloor\frac{n}{\text{lcm}(a,b)}\right\rfloor$ numbers that are divisible by both $a$ and $b$.

I have the intuition for this. E,g, let a=2, b=5 and n=200. Being divisible by 2 and 5 implies being divisible by 10, and 201 = 20*10 + 1 thus there are 20 numbers divisible by both 2 and 5.

However, I am not able to formulate a proof.

One of the first issues was when I was attempting to prove $a|x \cap b|x \implies \text{lcm}(a,b)|x$. I'd start with

$a|x \iff x = ac$ and $b|x \iff x = bd$ thus we have $x = ac = bd$ but not be sure where to continute. I know that $\text{lcm}(a,b) = \frac{ab}{\text{hcf}(a,b)}$ and think I need to apply this at some point but not sure how. Would this be easier to prove with the restriction that a,b are coprime?

Any help would be appreciated.

Two main steps to the proof... first prove that every number divisible by both $a$ and $b$ is divisible by $\text{lcm}(a,b)$. Next, prove that the number of numbers in the range $[1,n]$ who are divisible by $x$ is equal to $\left\lfloor \frac{n}{x}\right\rfloor$. The result follows. — JMoravitz, Sep 30 '20 at 11:41
Thank you. I am struggling with the first part, with my working included in the post. — hur, Sep 30 '20 at 11:44
Proving that any common multiple of two numbers is a multiple of their least common multiple. Also related to this fact but in the other direction: Concise proof that every common divisor divides gcd without bezouts identity — JMoravitz, Sep 30 '20 at 11:45

Proving there are $\left\lfloor\frac{n}{\text{lcm}(a,b)}\right\rfloor$ numbers divisible by a and b in [1,n]

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