(6a + 1) $\vdots$ (3a - 1)
$\Rightarrow$ (3a - 1) + ( 3a - 1) + 3 $\vdots$ ( 3a - 1)
$\Rightarrow$ 3 $\vdots$ (3a - 1)
$\Rightarrow$ 3a - 1 $\in$ Ư(3) = { $\pm1$; $\pm3$ }
Ta có:
3a - 1 = -3 $\Rightarrow$ a = $\frac{-2}{3}$ ( loại)
3a - 1 = -1 $\Rightarrow$ a = 0
3a - 1 = 1 $\Rightarrow$ a = $\frac{2}{3}$ (loại)
3a - 1 = 3 $\Rightarrow$ a = $\frac{4}{3}$ (loại)
Vậy a = 0.