I am trying to find a natural number $k>1$ such that $k$ divides $(26+35n)$ and $k$ divides $(3+7n)$ for some integer $n$. I know that $(ka)=(26+35n)$ for some $a \in Z$ and $(kb)=(3+7n)$ for some $b \in Z$ but I am not sure how to use to determine the value of $k$.

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    $\begingroup$ Calculate the gcd of both expressions $\endgroup$ – Rol Apr 10 '15 at 6:46

HINT: $k$ should divides $$(26+35n)-5(3+7n)=11$$ Also if you want to find the corresponding values of $n$ solve two linear Diophantine equations $$26+35n=11a\,\,\,\,\,\,\,\,\,\ 3+7n=11b$$ simultaneously for $a, b$ and $n.$

  • $\begingroup$ I understand that $k$ should divide $(26+35n)-5(3+7n)$ since both of those expressions are divisible by $k$. I am not sure I understand fully because then $k$ should divide $11$ which would mean $k=11$? $\endgroup$ – user3699546 Apr 10 '15 at 7:04

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