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I was trying to solve the problem in the title but couldn't. The solution was 1 or 13, but I couldn't go there. I am here:

Let's say d is gcd. Then $$d|5n+6,\quad d|7n-2\quad and\quad d|3*(7n-2) \quad lastly\quad d|5n+6+(3*(7n-2)) = d|26n $$ So ı can say $\quad d|26\quad or\quad d|n\quad$ but that's all.

ACrescendo
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    $7(5n+6)-5(7n-2) = 52$ do $d | 52$ for every $n$ – Will Jagy Oct 24 '24 at 23:23
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    You've done some good things here, but that final conclusion is wrong. If I tell you that $d \mid 26n$ for some integer $n$, then for the case $n = 2$, $d$ could be $4$, which is neither of the cases you cited. One basic rule is that if $p | ab$ and $p$ is prime, then $p | a$ or $p | b$. But in your case, you have no reason to believe that $d$ is prime, so you can't apply this rule. – John Hughes Oct 24 '24 at 23:24
  • Same methods in the dupe work here, e.g. mod reducing as in Euclid's algorithm yields the "remainder" sequence $,7n!-!2,,5n!+!6,,2n!-!8,,n!+!22,,52,$ so $,(7n!-!2,5n!+!6) = (n!+!22,52).\ \ $ – Bill Dubuque Oct 24 '24 at 23:53
  • Or in matrix form $\ \begin{bmatrix} 3&-2\ 7&-5\end{bmatrix} \begin{bmatrix} 5n+6\ 7n-2\end{bmatrix} = \begin{bmatrix} n+22\ 52\end{bmatrix}\ \ $ – Bill Dubuque Oct 25 '24 at 00:40
  • so you say d=1 and d=13 are wrong answers. – ACrescendo Oct 25 '24 at 05:41

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