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Basically, prove or disprove that $ab|c \implies a|c$.

It seems like it should, but I don't know how to go about proving it.

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    $\begingroup$ If $ab \mid c$ then we can write $c = abn$ for some integer $n$. Now let us add some suggestive parentheses: $c = a(bn)$. What can you conclude? $\endgroup$ – Bungo Mar 25 '16 at 1:53
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    $\begingroup$ just write down the definition of dividing. $\endgroup$ – Melquíades Ochoa Mar 25 '16 at 1:55
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ab|c means there exists an integer d such that (ab)d=c

So consider the integer bd now

a(bd)=c hence a divides c

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Hint: Doesn't "divides" really mean "is a factor of"? If so, then aren't factors of the divisor also factors of the dividend?

Example: $100$ is a factor of $1000$. Because $100=4\times 25$, you immediately know that $4$ is a factor of $1000$, and $25$ is a factor of $1000$.

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Hint:

Divisibility is a transitive relation.

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By definition, $ab|c \Leftrightarrow c = abk$ for some $k \in \mathbb{Z}$. Similarly, $a|c \Leftrightarrow c = am$ for some $m \in \mathbb{Z}$. Let $m = bk$ and you are done.

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