\(3n^2-13n+29=3n.\left(n-3\right)-4n+29\)
\(=3n.\left(n-3\right)-4.\left(n-3\right)+17=\left(3n-4\right).\left(n-3\right)+17\)
=> đề \(3n^2-13n+29⋮n-3\Rightarrow17⋮n-3\Rightarrow n-3\inƯ\left(17\right)=\left\{\pm1,\pm17\right\}\)
=> \(n\in\left\{4,2,-14,20\right\}\)
vì n là số nguyên dương => n\(\in\){4,2,20}