Finding a natural number $k>1$ such that $k$ divides $(26+35n)$ and $(3+7n)$

Question

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 $k$.

Answer 1

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 $n.$