Question

Let $P=\left\{1,2,3,4,8,16,.......1024\right\}$. Consider all possible positive differences of elements of $P$.Let $M$ is the sum of all these differences and $N$ be the sum of the digits of $M$ then

• A. $N$ is divisible by $3$ or $9$ or product of $3$ and $9$
• B. Number of prime factors of $N$ are $4$
• C. Sum of divisors of $N$ is $40$
• D. Number of divisors of $N$ are $4$

Answer 1

Answer: (A), (C), (D)

Given $P=\left\{2^0,2^1,2^2.....2^{10}\right\}$
$\therefore M=$Sum of all difference $=\sum _{s=1}^{10}\sum _{r=0}^{s-1}\left(2^s-2^r\right)$
$\therefore M=\sum _{s=1}^{10}\left[\left(2^s-2^0\right)+\left(2^s-2\right)+\left(2^s-2^2\right)+....+\left(2^s-2^{s-1}\right)\right]$
$=\sum _{s=1}^{10}\left(2^s+2^s+....\left(stimes\right)\right)-\left(2^0+2^1+......+2^{s-1}\right)]$
$=\sum _{s=1}^{10}\left[s{2}^s-2^0\frac{\left(2^s-1\right)}{2-1}\right]$
$=\sum _{s=1}^{10}\left[\left(s-1\right)2^s+1\right]$
$=\sum _{s=1}^{10}\left(s-1\right)2^s+\sum _{s=1}^{10}1$
$=0+4\left(1+2\cdot 2+3\cdot 2^2+....+9\cdot 2^8\right)+10$
$=4+8\cdot 2^{11}+10$ (using A.GP.)
$=14+2^{14}=14+\left(1024\right)16=16398$
$\therefore N=$ sum of digits of $M=1+6+3+9+8=27$
As $N=27$, which is divisible by $1,3,9$ and $27$ thus number of divisors of $N$ are $4$ and their sum is $1+3+9+27=40$