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So the goal is to prove that the $\operatorname{lcm}(a,b)$ divides any multiple of of $a$ and $b$. Suppose there is some integer $c$ such that $a|c$ and $b|c$ but I want to prove $\operatorname{lcm}(a,b)|c$ also. I got that $$\operatorname{lcm}(a,b)=\frac{(a\cdot b)}{\gcd(a,b)}$$ and I want to see how we could show $$\frac{(a*b)}{\gcd(a,b)}\bigg| \, c$$ Any help would be appreciated thank you.

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    $\begingroup$ It's the direction $(\Rightarrow)$ in the equiavlences in this answer in the first linked dupe. See the others for background and motivation.. $\endgroup$ Apr 27, 2020 at 18:08
  • $\begingroup$ Please don't change the question at this point. If you have a related question then pose a new question. $\endgroup$ Apr 28, 2020 at 1:18

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WARNING: This answer is INCORRECT. See comments for what mistakes I am making

Firstly we know that some multiple of $a$ and $b$ can be represented by $kab$ where $k$ is some integer, and thus is divisible by $ab$.

We need to show that $lcm(a,b) \vert ab$. Since $gcd(a,b)$ would be some integer, and that $lcm(a,b) gcd(a,b) = ab$, it is obvious.

Therefore any multiple of $ab$ is divisible by $lcm(a,b)$.

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    $\begingroup$ The claim in the first sentence is false. $\endgroup$ Apr 27, 2020 at 18:12
  • $\begingroup$ I wonder, why it is not true that $ a \vert c $ and $ b \vert c $ means $ (ab) \vert c$ and therefore $c = kab$ ? (I know that maybe gcd(a,b) is not 1 but c is some multiple of a and b) $\endgroup$ Apr 27, 2020 at 22:39
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    $\begingroup$ e.g. $\,2,4\mid 4\,$ but not $\,2\cdot 4\mid 4.\ $ Your claim is true $\iff \gcd(a,b) = 1\ \ $ $\endgroup$ Apr 27, 2020 at 22:44
  • $\begingroup$ Thanks. I thought that multiple of a and b means that $c = k a ^m b^n$ where k,m,n are some integers...... $\endgroup$ Apr 27, 2020 at 22:52

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