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Prove that if $\gcd(a,c)=1$ and $\gcd(b,c)=1$ then $\gcd(ab,c)=1$. My attempt is let $\gcd(ab,c)=d$. Since $d \mid ab$ and $d \mid c$ , $d \mid (abt+cs)$ for some integers $s$ and $t$. Then by definition of divisibility, we have $$abt+cs=dk$$ for some integers $k$. Then I got stuck here.