Có $\left( 3n+8 \right)\vdots \left( 2n+1 \right)$
Thì $2\left( 3n+8 \right)\vdots \left( 2n+1 \right)$
Hay $2\left( 3n+8 \right)-3\left( 2n+1 \right)\vdots \left( 2n+1 \right)$
Suy ra: $13\vdots \left( 2n+1 \right)$
Hay $\left( 2n+1 \right)\in U\left( 13 \right)=\left\{ 1;13 \right\}$
+) TH1: $2n+1=1$
Suy ra $n=0.$
+) TH2: $2n+1=13$
Suy ra $n=6.$
Vậy $n\in \left\{ 0;6 \right\}$