Biết 3n + 10 chia hết cho 2n + 1. Tìm n.

Ta có: \[(3n + 10)\,\, \vdots \,\,(2n + 1)\]

\[ \Rightarrow [2(3n + 10)]\,\, \vdots \,\,(2n + 1)\]

\[ \Rightarrow (6n + 20)\,\, \vdots \,\,(2n + 1)\]

\[ \Rightarrow (6n + 3 + 17)\,\, \vdots \,\,(2n + 1)\]

\[ \Rightarrow [3(2n + 1) + 17)]\,\, \vdots \,\,(2n + 1)\]

Vì \[[3(2n + 1)]\,\, \vdots \,\,(2n + 1)\] nên \[17\,\, \vdots \,\,(2n + 1)\]

Þ (2n + 1) ∈ Ư(17)

Þ (2n + 1) = { -17; -1; 1; 17}

Þ n = {-9; -1; 0; 8}

Vậy n = {-9; -1; 0; 8}