How many different numbers which are smallerr than `2xx10^(8)` and are divisible by 3, can be written by means of the digits 0,1 and 2?

Explanation

12,21 . . . 122222222 are form the required numbers we can assume all of them to be nine digit in the form.
`a_(1),a_(2),a_(3),a_(4),a_(5),a_(6),a_(7),a_(8),a_(9)` and can use 0 for `a_(1),a_(2)` and `a_(0)` and `a_(0),a_(1),a_(2)` and `a_(3)` . . and so on to get 8-digit, 7-digit, 6digit numbers etc. `a_(1)` can assume one of the 2 values of 0 or 1. `a_(2),a_(3),a_(4),a_(5),a_(6),a_(7),a_(8)` can assume any of 3 values 0,1,2.
the number for which
`a_(1)=a_(2)=a_(3)=a_(4)=a_(5)=a_(6)=a_(7)=a_(8)=a_(9)=0` must be eleminated. the sum of first 8-digits i.e., `a_(1)+a_(2)+ . . .+a_(8)` can be in the form of `3n-2` or 3n-1 or 3n.
in each case `a_(9)` can be chosen from 0,1,2 in only 1 way so that the sum of all 9-digits in equal to 3n.
`therefore`Total numbers=`2xx3^(7)xx1-1=4374-1=4373`.