Question

If $S$ is the sum of all digits of the number $N$ where $N=1+11+111+.....$ upto $2011$ terms then $S$ is a/an

• A. perfect square of an odd number
• B. prime number
• C. perfect cube
• D. square of even number

Answer 1

Answer: (B)

$N=1+11+111+.....$ to $2011$ terms
$=\frac{1}{9}\left[\left(10-1\right)+\left(10^2-1\right)+.....to2011terms\right]$
$=\frac{1}{9}\left[10+10^2+....+10^{2011}-2011\right]$
$=\frac{1}{9}\left[\frac{10\left(10^{2011}-1\right)}{9}-2011\right]$
$=\frac{1}{81}\left[10^{2011}-18109\right]$
$=\frac{1}{81}\left[\mathrm{999......981891}\right]$ begins with $\left(2012-m\right)$ nines
where $m=$ number of digits of $18109$ i.e. $m=5$
$\therefore N=\frac{1}{9}\left[\mathrm{111....109099}\right]$ begins with$\left(2012-5=2007\right)$ ones
$N=12345679$ (each digit $223$ times) $01011$
$\left(\because \frac{2007}{9}=223\right)$
$=223\times \left(1+2+3+4+5+6+7+8+9\right)+1+0+1+1$
$=223\times 37+3=8251+3=8254$
$\therefore S=8+2+5+4+19$ which is a prime number