**and**$x$ is divisible by 3?

Yes, if a number $n$ is divisible through $15$, this means $n=15k$ for some integer $k$. For this reason $n=5(3k)=3(5k)$, so the is likewise divisible by $3$ and $5$.

Conversely, if $n$ is divisible through $3$ and also $5$, that is a straightforward lemma that it is divisible through the least common multiple that $3$ and also $5$. Due to the fact that $3$ and $5$ space coprime, their lcm is simply $15$.

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$egingroup$ ns feel that this price is incomplete. That does no prove that x|n and y|n indicates that lcm(x,y)|n; it just insurance claims it's "a simple lemma". A much better answer would display why this is true. $endgroup$

$egingroup$ expect there's a number n such that x|n and y|n, but lcm(x,y) does no divide n. Then compose m = lcm(x,y) and n = pm+q, whereby 0 $endgroup$

Yes. $15|n implies 3*5|n implies 3|n ext and also 5|n$. Vice versa, $3|n ext and also 5|n implies 15|n$. This is due to the fact that if $x|n$ and also $y|n$, then $ extlcm(x,y)|n$ whereby $ extlcm(xy)$ is the smallest number the is divisible through both $x$ and also $y$. In this case, $x=3$ and $y=5$, therefore $ extlcm(3,5) = 15$, as such $15|n$.

(Note: $a|b$, review as "$a$ divides $b$", means that $b$ is divisible by $a$.)

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