What is the converse of this statement and is it true? If $a$ and $b$ are relatively prime, $a\mid c$ and $b\mid c$, then $(ab)\mid c$.
I am lost. Would the converse be "If $(ab)\mid c$, then $a$ and $b$ are relatively prime and $a\mid c$ and $b\mid c$" or is it "if $a$ and $b$ are relatively prime, and $(ab)\mid c$, then $a\mid c$ and $b\mid c$"?
It seems like the second variation would be more fitting, but I'm not sure. 
 A: The word "converse", as it is practically used in mathematics text if not necessarily by dictionaries of logic, is somewhat fuzzy, and the meaning of "the converse of theorem such-and-such" sometimes has to be deduced from the context.
As long as we only have atomic claims $P$ and $Q$ with the implication $P\to Q$, then without doubt the converse of $P\to Q$ is $Q\to P$. But when there are more than one premise, some room for interpretation opens up.
The problem is that in the usual style of written mathematics, the two theorems

Theorem 1. If $a$ and $b$ are coprime and $a\mid c$ and $b\mid c$, then $ab\mid c$.

and

Theorem 2. Assume that $a$ and $b$ are coprime. If $a\mid c$ and $b\mid c$, then $ab\mid c$.

mean exactly the same thing -- usually we're not even trained to notice the difference between them as we read a mathematical text. However, according to a strict logical interpretation of "converse" these two clearly equivalent theorems would have different converses:

Converse 1. If $ab\mid c$, then $a$ and $b$ are coprime and $a\mid c$ and $b \mid c$.
Converse 2. Assume that $a$ and $b$ are coprime. If $ab\mid c$, then $a\mid c$ and $b\mid c$.

In practice, however, most authors don't care about this, and just speak of "the converse" with the meaning "the possible interpretation of converse that makes sense in the context". The reader is supposed to figure out for himself which of the premises look like something that could conceivably have a reasonable chance of being consequences of the original conclusion, given the other premises.

In the language of formal logic, another way to express the problem is that ordinary mathematical reasoning (presented in natural language) doesn't distinguish consistently between


*

*$P_1 \vdash P_2\to Q$

*$P_1 \to (P_2 \to Q)$

*$(P_1 \land P_2) \to Q$


These are generally just different formal representations of the same concept inside the working mathematician's mind, and it can take some training and experience with formal logic to even appreciate that a useful distinction between them can be made. Therefore a notion of "converse" that assigns crucial meaning to these essentially syntactic differences will have a hard time agreeing with how the word is used in mathematical writing that is not concerned with logic in particular.
A: Your first interpretation is the correct interpretation.
You have a statement, roughly speaking, consisting of the form $$(p\land q \land r) \rightarrow s$$
where $p$ denotes $\gcd(a, b) = 1$, $\;q$ denotes $a \mid c$, $\;r$ denotes $b \mid c$, and $\;s$ denotes $(ab) \mid c$.
The converse of that implication is $$s \rightarrow (p \land q \land r)$$

Put more generally, the converse of any implication "if P, then Q" is given by "if Q, then P".
In your case, $P$ happens to be:  "$a$ and $b$ are relatively prime and $a\mid c$ and $b \mid c$"
whereas $Q$ is given by "$(ab)\mid c$".
As to whether the converse is true?:
No the converse is not true. Let $a = 2, b = 4, c = 16.$ Then $(ab) = 8 \mid 16 = 2$, but $\gcd(a, b) =\gcd(2, 4) = 2 \neq 1$: i.e., $a = 2$ and $b = 4$ are not relatively prime. Hence, the converse is not true for all integers $a, b, c, \;c\neq 0$
