Question

Let $S\left(n\right)$ denote the sum of the digits of a positive integer $n$. e.g. $S\left(178\right)=1+7+8=16$. Then, the value of $S\left(1\right)+S\left(2\right)+S\left(3\right)+\cdots +S\left(99\right)$ is

• A. $476$
• B. $998$
• C. $782$
• D. $900$
• E. $855$

Answer 1

Answer: (C)

$782$

$S\left(1\right)+S\left(2\right)+S\left(3\right)+\cdots +S\left(99\right)=$ Sum of digits of counting numbers $1$ to $99$.
We know that, in $1$ to $99$, there is $20$ times $1$, $20$ times $2$, $20$ times $3$, $\dots ,20$ times $9$ and $9$ times zero.
So, $S\left(1\right)+S\left(2\right)+S\left(3\right)+\cdots +S\left(99\right)$
$=20\times 1+20\times 2+20\times 3+\cdots +20\times 9+9\times 0$
$=20[1+2+3+\cdots +9]$
$=\frac{20\times 9\times (9+1)}{2}$ $[\because \sum _{n=1}^{n}n=\frac{n(n+1)}{2}]$
$=\frac{20\times 9\times 10}{2}=9\times 10\times 10=900$