Question

Let $k$ be the number of different numbers which are smaller than $2\times {10}^8$ and are divisible by $3$, can be written by means of the digits $0,1$ and $2$. Find sum of digits of $k$ ?

Answer 1

$12,21,....,122222222$ are the required numbers we can assume all of them to be nine-digit in the form of $a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,a_9$ and can use $0$ for $a_1;a_2$ and $a_0$ and $a_0,a_1,a_2,...$ and so on to get $8$ digit, 7, digit 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 the three 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 eliminated. The sum of first $8$ digits i.e. $a_1+a_2+a_3+a_4+a_5+a_6+a_7+a_8$ 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 is equal to $3n$.
$\therefore$ Total numbers $=2\times 3^7\times 1-1=4374-1=4373$
Sum of digits is 17