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Let $m$ and $n>1$ two integers. Find sufficient and necessary conditions in which $\frac{6m+2}{n}$ is not a natural number.

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$6m+2$ is always congruent to $2\pmod 3$, so $n=3$ will always result in a non natural number. So it is sufficient that $n=3$, but not necessary since $n=4$ and $m=4$ yields a non-natural number.

I think the most succinct way to put this is that you need $6m+2\equiv0\pmod n$. I am not certain much more can be said without further restrictions on either $m$ or $n$.

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  • $\begingroup$ @ Laars Helenius: Can you elaborate plaese. $\endgroup$
    – DER
    Nov 1, 2014 at 16:08
  • $\begingroup$ See my edits above. I would appreciate the down vote (whomever is responsible) to reconsider my edits as well. Thanks. $\endgroup$ Nov 1, 2014 at 18:43

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