Find all the numbers such that $\displaystyle\frac{2n-7}{5n-23}$ can't be reduced

My attempt:

We want that:$\quad\gcd(2n-7,5n-23)=1$

that's equal to $\quad\gcd(2n-7,3n-16)\\ \quad=\gcd(2n-7,n-9)\\ \quad=\gcd(2n-7,11)$

So the answer is when $11\nmid 2n-7$

I am not sure about what I did, is it correct?

  • 2
    $\begingroup$ Yep, it is correct. $\endgroup$ – S.C.B. May 29 '16 at 15:57
  • 4
    $\begingroup$ You may be expected to say that it can be reduced iff $2n\equiv 7\pmod{11}$, or equivalently $n\equiv 9\pmod{11}$, so can't be reduced iff $n\not\equiv 9\pmod{11}$. $\endgroup$ – André Nicolas May 29 '16 at 16:02
  • $\begingroup$ Yes, thank you @AndréNicolas $\endgroup$ – Error 404 May 29 '16 at 16:03
  • $\begingroup$ Another way is to notice that $5(2n-7)-2(5n-23)=11$, so $\gcd(2n-7,5n-23)\mid 11$, so $\gcd(2n-7,5n-23)\in\{1,11\}$, so we have the equivalence $\gcd(2n-7,5n-23)=1\iff 11\nmid 2n-7$ (or equivalently $\gcd(2n-7,5n-23)=1\iff 11\nmid 5n-23$). $\endgroup$ – user236182 May 29 '16 at 17:33

We have \begin{align*}\gcd(2n-7,5n-23) &= \gcd(2n-7,n-9)\\ &= \gcd(11,n-9) = 1.\end{align*} Thus, $n \not \equiv 9 \pmod{11}$.


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