Ta có: \(6n+1⋮3n-1\)
\(\Leftrightarrow1⋮3n-1\)
\(\Leftrightarrow3n-1\inƯ\left(1\right)\)
\(\Leftrightarrow3n-1\in\left\{1;-1\right\}\)
\(\Leftrightarrow3n\in\left\{2;0\right\}\)
\(\Leftrightarrow n\in\left\{\frac{2}{3};0\right\}\)
Vì \(n\in Z\) nên n=0
Vậy:khi n=0 thì \(6n+1⋮3n-1\)