The total number of 4-digit numbers whose greatest common divisor with 18 is 3, is

$$\begin{gathered} \mathrm{Let} \mathrm{x} \mathrm{be} \mathrm{the} 4 \mathrm{digit} \mathrm{number} \\ \mathrm{such} \mathrm{that} \gcd (x, 18)=3 \\ \therefore \mathrm{x} \mathrm{is} \mathrm{odd} \mathrm{integer} \mathrm{which} \mathrm{is} \mathrm{divisible} \mathrm{by} 3 \mathrm{but} \mathrm{not} \mathrm{by} 9. \\ 4 \mathrm{digit} \mathrm{odd} \mathrm{numbers} \mathrm{which} \mathrm{are} \mathrm{multiples} \mathrm{of} 3 \\ \mathrm{A}=\left\{1005, 10011, 1017, \dots ,9999\right\} \\ {\mathrm{t}}_{\mathrm{n}}=9999=1005+(\mathrm{n}-1)6\Rightarrow \mathrm{n}=1500 \\ 4 \mathrm{digit} \mathrm{numbers} \mathrm{odd} \mathrm{multiples} \mathrm{of} 9 \\ \mathrm{B}=\left\{1017,1035,\dots ,9999\right\} \\ 9999=1017+(\mathrm{n}-1)18\Rightarrow \boxed{\mathrm{n}=500} \\ \therefore \mathrm{Required} \mathrm{number} \mathrm{of} 4 \mathrm{digit} \mathrm{numbers} \\ =\mathrm{n}\left(\mathrm{A}\right)-\mathrm{n}\left(\mathrm{B}\right) \\ =1500-500=1000 \end{gathered}$$