Is stating that a number $x$ is divisible by 15 the same as stating that $x$ is divisible by 5 and $x$ is divisible by 3?

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    $\begingroup$ The two assertions are equivalent. $\endgroup$ – André Nicolas Feb 1 '12 at 23:38
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    $\begingroup$ may be worth noting for somebody who might be new to this, that it happens to work in both directions here, because 3 and 5 are co-prime. i.e. if you ask the same question for a different example you may not get the converse implication. $\endgroup$ – Beltrame Feb 1 '12 at 23:49
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    $\begingroup$ If you want to figure out the general facts at work here, you could read this Keith Conrad handout. $\endgroup$ – Dylan Moreland Feb 1 '12 at 23:55

Yes, if a number $n$ is divisible by $15$, this means $n=15k$ for some integer $k$. So $n=5(3k)=3(5k)$, so it is also divisible by $3$ and $5$.

Conversely, if $n$ is divisible by $3$ and $5$, it is a simple lemma that it is divisible by the least common multiple of $3$ and $5$. Since $3$ and $5$ are coprime, their lcm is just $15$.

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  • $\begingroup$ I feel that this answer is incomplete. It does not prove that x|n and y|n implies that lcm(x,y)|n; it just claims it's "a simple lemma". A better answer would show why this is true. $\endgroup$ – user22805 Feb 2 '12 at 8:32
  • $\begingroup$ Suppose there's a number n such that x|n and y|n, but lcm(x,y) does not divide n. Then write m = lcm(x,y) and n = pm+q, where 0 < q < m and p is an integer. Then x and y must both divide q, so m is not the lcm of x and y - contradiction. $\endgroup$ – user22805 Feb 2 '12 at 8:35

Yes. $15|n \implies 3*5|n \implies 3|n \text{ and } 5|n$. Conversely, $3|n \text{ and } 5|n \implies 15|n$. This is because if $x|n$ and $y|n$, then $\text{lcm}(x,y)|n$ where $\text{lcm}(xy)$ is the smallest number that is divisible by both $x$ and $y$. In this case, $x=3$ and $y=5$, so $\text{lcm}(3,5) = 15$, therefore $15|n$.

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

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    $\begingroup$ You should also prove the other direction as Dedede has done below. $\endgroup$ – user38268 Feb 1 '12 at 23:51
  • $\begingroup$ @BenjaminLim: Good point. Editing that in now. $\endgroup$ – El'endia Starman Feb 1 '12 at 23:53
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    $\begingroup$ @El'endiaStarman: You seem to have done nothing except write the OP's question as an assertion using mathematical language. A proof or justification as to why this works would be helpful. $\endgroup$ – JavaMan Feb 1 '12 at 23:59
  • $\begingroup$ I don't know enough math to pick the "most correct" answer. @El'endiaStarman answered first, I understood it, and it looked good to me! Dedede's has more upvotes though, so I went with that one. $\endgroup$ – Adam Monsen Feb 2 '12 at 0:02
  • $\begingroup$ @JavaMan: At this point, is there a reason I shouldn't just delete my answer? Dedede has already answered it perfectly. $\endgroup$ – El'endia Starman Feb 2 '12 at 0:02

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