Question

The sum of all three-digit natural numbers which leave a remainder $2$ when divided by $3$

• A. $168450$
• B. $168850$
• C. $165840$
• D. None of these

Answer 1

The correct option is A $168450$

First three digit number leaves remainder $2$ is $101$

The next being $104,107,110$,...…

Last $3$ digit number $=998$

$\therefore 101,104,\dots \dots 998$

$a=101,d=3$

$n^{th}$ term $\Rightarrow 998=a+(n-1)d$

$998=101+(n-1)3$

$(n-1)\text{/}3={\text{/}897}_{299}$

$n=300$.

$S_n=\frac{300}{2}\left(2\right(101)+(300-1\left)3\right)=\frac{300}{2}(202+897)$

$=164850$.