Đáp án: $n\in\{13k+7,13q-3\}, k,q\in N$
Giải thích các bước giải:
Ta có:
$2n^2+18n-3\quad\vdots\quad 13$
$\to 2n^2+13n+5n-3\quad\vdots\quad 13$
$\to 2n^2+5n-3\quad\vdots\quad 13$
$\to 2n^2+6n-n-3\quad\vdots\quad 13$
$\to (2n^2+6n)-(n+3)\quad\vdots\quad 13$
$\to 2n(n+3)-(n+3)\quad\vdots\quad 13$
$\to (2n-1)(n+3)\quad\vdots\quad 13$
$\to 2n-1\quad\vdots\quad 13$ hoặc $n+3\quad\vdots\quad 13$
Nếu $2n-1\quad\vdots\quad 13$
$\to 2n-1=13(2k+1), k\in N$ vì $2n-1$ lẻ
$\to 2n-1=26k+13$
$\to 2n=26k+14$
$\to n=13k+7$
Nếu $n+3\quad\vdots\quad 13$
$\to n+3=13q\to n=13q-3, q\in N$
