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Let $n, m$ and $p$ non-zero natural integers, with $n$ and $p$ relatively prime. Prove that $\gcd(n, mp) = \gcd (n, m)$.

This problem had three questions. First, to prove that if $d$ divides $n$ then $d$ and $p$ are relatively prime. That's done. Second, to prove that an integer $d$ which divides $n$ and $mp$ also divides $m$. Done. I'm left with the third question (the one in the title). I know I'm supposed to use the results I got for the first two, but I just can't seem to connect the dots...

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    So, you have proved that if $d=\gcd(n,mp)$, then $d$ divides $m$, so $d$ is a common divisor of $m$ and $n$. There's just a little bit left to do....2012-12-11

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