Tìm số tự nhiên n sao cho (5n + 2) chia hết cho (9 – 2n)
Ta có: $\left( 5\text{x}+2 \right)\,\,\vdots \,\,\left( 9-2n \right)$
$\Rightarrow 2\left( 5n+2 \right)\,\,\vdots \,\,\left( 9-2n \right)\,\,$
$\Rightarrow \left( 10n+4 \right)\,\,\vdots \,\,\left( 9-2n \right)$
$\Rightarrow \left( -45+10n+45+4 \right)\,\,\vdots \,\,\left( 9-2n \right)$
$\Rightarrow -5\left( 9-2n \right)+49\,\,\vdots \,\,\left( 9-2n \right)$
Vì $-5\left( 9-2n \right)\,\,\vdots \,\,\left( 9-2n \right)$
$\Rightarrow 49\,\,\vdots \,\,\left( 9-2n \right)$ hay $\left( 9-2n \right)$ ∈ Ư(49) = {1; 7; 49}
9 – 2n 1 7 49
n 4(TM) 1(TM) (KTM)
Vậy n ∈ {1;4}