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I'm not going to list the choices here, mainly because I just want the general idea. If I generalize the question and was given $n$ different integers divide some integer $r$, how do I determine what else it is divisible by?

My initial attempt at the problem was that it should be divisible by $2$ (since $8$ is divisible by $2$), and $3$ and $5$ (since $15$ is divisible by those). Then I thought the number should be divisible by $2, 3,$ and $5$. But apparently that's incorrect.

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up vote 2 down vote accepted

You are on the right track, but being divisible by $8$ implies more than being divisible by $2$.

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What else does it imply? (I actually thought about it, but I'm quite hesitant to state what I think). – AlanH Dec 20 '12 at 6:14
@AlanH.: All numbers divisible by $8$ are divisible by $2$, but not the the other way. If you know it is divisible by $8$ and $15$, you know it is divisible by $lcm (8,15)=120$ – Ross Millikan Dec 20 '12 at 6:21
Ah! I figured it out just now. Thank you – AlanH Dec 20 '12 at 6:25

In general, if an integer is divisible by $a_1,a_2,\ldots,a_k$, then the integer is divisible by $\text{lcm}(a_1,a_2,\ldots,a_k)$. The proof of this claim immediately follows from the fact that if $a \vert bc$ and $\gcd(a,b) = 1$, then $a \vert c$, which follows immediately from the definition of $\gcd$.

Hence, in your case, the integer is divisible by $120$.

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If $a$ is divisible by $b$, then $a$ is also divisible by all the factors of $b$. If $a$ is divisible by $b$ and $c$, then $a$ is divisible by products of the factors of $b$ and $c$. Does that help?

You could also calculate the smallest number that is divisible by both $8$ and $15$ and find what numbers divide that number.

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If $a$ is divisible by $b$ and $c$, then $a$ is divisible by products of the factors of $b$ and $c$. Well not exactly: $30$ is divisible by $6$ and by $10$, but is not divisible by $60$. – ShreevatsaR Dec 20 '12 at 14:23
I didn't say all of the products. – Todd Wilcox Dec 20 '12 at 14:23
That's my point, that it's not clear what you meant. For instance one could say: $1$ is always a factor of $b$ and $c$, so $a$ is divisible by the product $1 \times 1 = 1$. But this doesn't help. – ShreevatsaR Dec 20 '12 at 14:37
Duly noted. I tend to break things down and over-simplify since I'm used to tutoring High School students. Based on the complexity level of the original question and the presence of the homework tag, I felt that it was reasonable to present a non-rigorous but more understandable answer. – Todd Wilcox Dec 20 '12 at 15:03

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