# Prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$. [duplicate]

Let $a,b,c \in \mathbb{Z}$, prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$.

## marked as duplicate by Martin Sleziak, mrp, user133281, Namaste, LeucippusOct 24 '16 at 1:40

• It would be useful if you could tell us what you have tried. This helps us to understand what you already know so we can better focus our help. Also, people around here get touchy when people just post a question without motivation etc. as they are unwilling to simply do others homework for them, moreover they dislike it when a third party steps in and does the homework for them anyway so questions like this which merely states a question tend to get closed (pending adding more info). Adding in what you have tried and where you came across it will stop your question getting closed in this way. – user1729 Oct 22 '13 at 19:08

Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively

Clearly, at least one of $A,B$ is zero,

Let $A=0,B\ge0$

The highest power of prime $p$ in $(ab,c)=$min$(A+B,C)=$min$(B,C)$

The highest power of prime $p$ in $(a,c)=$min$(A,C)=0$

The highest power of prime $p$ in $(b,c)=$min$(B,C)$

The highest power of prime $p$ in $(b,c)\cdot(a,c)=$min$(B,C)+0$

This holds true for all prime that divides $c$