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If $a, b, c$ and $k$ be integers, $\gcd(a,b) = 1$ and $\gcd(a, c)=k$, then $\gcd (bc, a)=k$.

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    Yes, that is true. Did you have a question though?2011-09-17
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    See [this](http://math.stackexchange.com/questions/62072/how-to-show-that-gcdab-n-1/62078#62078). Same idea works; or, replace $a$ and $c$ with $a/k$, $c/k$; then $\gcd(a/k,b) |\gcd(a,b) = 1$, so $\gcd(a/k,b)=1$, and you reduce to that case.2011-09-17
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    $k = 1\times k = (a,b)(a,c) = (a(a,b,c),bc) = (a,bc)$.2011-09-17
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    Someone has voted to close this question as a duplicate of [another question](http://math.stackexchange.com/questions/62072/how-to-show-that-gcdab-n-1). But I do not see how this is a duplicate.2011-09-18

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