$(ab,c) = (a,c)(b,c)$ (weakest condition possible on $a,b,c$) Let $(x,y)$ be the greatest common divisor of integers, $x,y$. The way I tried to approach the problem in the title is first we can prove that if $(a,c) = 1$ or $(b,c) = 1$ then that equality holds. That is $(ab,c) = (a,c)(b,c)$. I am trying to generalize it and get the weakest condition one could get to ensure the equality.
I was thinking of it as follows we know that with
$$a = p_1^{a_1}\ldots p_n^{a_n}$$
$$b = p_1^{b_1}\ldots p_n^{b_n}$$
we have $(a,b) = p_1^{q_1} \ldots p_n^{q_n}$, where $q_i = \min(a_i,b_i)$
So in general what we are trying to find is the following when exactly does the following condition occurs:
$\min(a_i + b_i,c_i) = \min(a_i,c_i) + \min(b_i,c_i)$, but at this point I didn't know how to proceed further.
I know that if $(a,c) = 1$ then we get an answer however is that the weakest condition possible to get $(ab,c) = (a,c)(b,c)$?
 A: Write $\ A,B,C = ad,bd,cd,\,\ d = (A,B,C),\, $ so $\,(a,b,c) = 1,\,$ for $A,B,C\,$ not all $\,0,\,$ so $d\neq 0$.
Then $\,(AB,C) = (abd^2,cd) = d(abd,c)$
and $\ (A,C)(B,C) = d^2(a,c)(b,c) = d^2(ab,c(a,b,c)) = d^2(ab,c)$
The two are equal iff $\ (abd,c) = (abd,cd),\, $ e.g. if $\ d = 1.\,$ But for any $\,d>1\,$ it can fail, e.g. $\, a = 1,\ b = c = d\,\Rightarrow\, (abd,c)=(d^2,d)=d\,$ vs. $(abd,cd) = (d^2,d^2) = d^2$
A: First of all: Think that $4=(4.4,4)$ and $(4,4)(4,4)=4.4=16$ So we saw that if condition not hold then it can be false.
Now assume $(a,c)=1$ and say $(ab,c)=d$ (Note that: naturally $d|c$ and (b,c)|d). There exist $x,y \in Z$ such that $ax+cy=1$. 
Now multiply with $b$: $(ab)x+(c)by=b$. This means $d|b$. Thus $d|(b,c)$ and hence $d=(b,c)$.
A: The formula $$\gcd(ac,bc)=\gcd(a,c)\cdot\gcd(b,c)\tag{1}$$
defines a $3$-ary relation among natural numbers, in other words: a certain subset $R\subset{\mathbb N}^3$. There is no "weakest condition" for this relation. For any triple $(a,b,c)$ either $(a,b,c)\in R$ or $(a,b,c)\notin R$.
One may ask however, whether the definition of $R$ can be simplified, i.e., be written with less symbols. You have made a first step in this direction; whether it leads to a simplification is debatable.
For  any given prime $p$, put $\alpha:=\max\{ k|\ p^k|a\}$, and define $\beta$, $\gamma$ similarly. Then $(1)$ is true iff
$$\min\{\alpha+\beta,\gamma\}=\min\{\alpha,\gamma\}+\min\{\beta,\gamma\}\tag{2}$$
holds for all $p$. (That's how far you got.)
We now have to analyze $(2)$. When $\alpha=0$ then $(2)$ is trivially true. Assume now that $0<\alpha\leq \beta$, and keep $\alpha$ and $\beta$ fixed. Then
$$\min\{\alpha+\beta,\gamma\}=\cases{\gamma &$(\gamma\leq\alpha+\beta)$ \cr \alpha+\beta&$(\gamma\geq\alpha+\beta)$\cr}$$
and
$$\min\{\alpha,\gamma\}+\min\{\beta,\gamma\}=\cases{ 2\gamma&$(\gamma\leq\alpha)$ \cr \alpha+\gamma&$(\alpha\leq\gamma\leq\beta)$\cr
\alpha+\beta&$(\gamma\geq\beta)\ .$\cr}$$
From this it follows that $(2)$ is violated iff $0<\gamma<\alpha+\beta$ (draw the graphs!). As a conclusion we can say that $(2)$ is equivalent with
$$\min\{\alpha,\beta,\gamma\}>0\quad\Longrightarrow\quad \alpha+\beta\leq\gamma\ .$$
