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I am really struggling with this problem please help.

3 Answers 3

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We need $(7n+1,11n+7)>1$

Now if integer $d$ divides $7n+1,11n+7; d$ must divide $7(11n+7)-11(7n+1)=38$

So, the necessary condition both $11n+7,7n+1$ are divisible by $2$ or $19$ or $38$

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    No. $11n+7$ does not necessarily divide $38$. Rather, $11n+7$ must be a multiple of $d$ where $d$ divides $38$ and is greater than one. I can edit but prefer to let you correct. – 2017-02-21
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    @OscarLanzi, Thanks. Please find the updated answer – 2017-02-21
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A sufficient condition for it to not be in lowest terms is that $n$ is odd.

To prove this, notice that, if $n$ is odd, then $7n+1$ is even and $11n+7$ is even as well. Therefore, both numerator and denominator can be divided by $2$.

Hope this partial answer helps.

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The Euclidean algorithm gives $$ \gcd(7n+1,11n+7)=\gcd(n+11,38) $$ Therefore $\gcd(7n+1,11n+7)>1$ iff $n+11$ is a multiple of $2$ or a multiple $19$.

In other words, the fraction is in lowest terms iff neither $2$ nor $19$ divide $n+11$.

This reduces to $n$ even but not of the form $38k+8$.