The number of 4-digil numbers whose GCD with 18 is 3, is

We find that: The number of 4 digit numbers = 9 x 10 x 10 x 10 = 9000 These 9000 numbers can be divided into 9 groups each consisting of 1000 numbers. Group $G_i$contains numbers of the for $9k+i,i=0,1,2,\dots .8$. The GCD of any number out of 9000 numbers with 18 will be three if the number is odd and multiple of 3 but not that of 9. So, the number should be of the form $9k + 3.$ The total number of such numbers is 1000.