Question

Let $x_1,x_2,...,x_k$ be the divisors of positive integer ${}^{\prime }{n}^{\prime }$ (including $1$ and $n$). If $x_1+x_2+...+x_k=75$, then $\sum _{i=1}1/x_i$ is equal to

• A. $\frac{75}{n^2}$
• B. $\frac{75}{n}$
• C. $\frac{75}{k}$
• D. None of these

Answer 1

The correct option is B $\frac{75}{n}$

If $x_1,x_2,.........x_k$ are divisors of ${}^{\prime }{n}^{\prime }$ in ascending order then $x_1\times x_k=n,x_2\times x_{k-1}=n$ and so on [ For eq. $:12,-1,2,3,4,5,6,12,$ $1\times 12=12,2\times 6=12,3\times 4=12$ ]

Given that $x_1+x_2+.........x_k=75$

$\Rightarrow {\sum }_{i=1}^k\frac{1}{x_i}=(\frac{1}{x_1}+\frac{1}{x_2}+......+\frac{1}{x_k})\times \frac{n}{n}$

$=\frac{1}{n}(\frac{n}{x_1}+\frac{n}{x_2}+......+\frac{n}{x_k})$

$=\frac{1}{n}(x_k+x_{k-1}+.......+x_k)$ ( From earlier result )

$=\frac{1}{n}\times 75$

$\therefore {\sum }_{i=1}^k\frac{1}{x_i}=\frac{75}{n}$

Hence, the answer is $\frac{75}{n}.$