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
$= \frac{1}{9} \left[\left(10 -1\right) +\left(10^{2} -1\right)+ ..... to 2011 terms\right]$
$= \frac{1}{9} \left[10 +10^{2} +....+10^{2011}-2011\right]$
$= \frac{1}{9}\left[\frac{10\left(10^{2011}-1\right)}{9}-2011\right]$
$= \frac{1}{81}\left[10^{2011} -18109\right]$
$= \frac{1}{81}\left[999 ......981891\right]$ begins with $\left(2012 - m\right)$ nines
where $m =$ number of digits of $18109$ i.e. $m = 5$
$\therefore N = \frac{1}{9}\left[111 ....109099\right]$ 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