If x is a digit such that the number

It is given that $\overline{18 \times 71}$ is a multiple of 3 .

$\therefore(1+8+\mathrm{x}+7+1)$ is a multiple of 3 .

$\therefore(17+x)$ is a multiple of 3 .

$\therefore 17+x=0,3,6,9,12,15,18,21 \ldots$

But $x$ is a digit. So, $\mathrm{x}$ can take values $0,1,2,3,4 \ldots 9$.

$17+x=18 \Rightarrow \mathrm{x}=1$

$17+x=21 \Rightarrow \mathrm{x}=4$

$17+x=24 \Rightarrow \mathrm{x}=7$

$x=1,4,7$