Question

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

To find: ${\sum }_{i=1}^n{\frac{1}{x}}_i$

Multiply and divide by $n$

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

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

Now, as $x_1$ is divisor of $n$, so $\frac{n}{x_1}$ is also a divisor and it's value will be either $x_1$ or $x_2$ or ... $x_n$.

For each $x_i$ term we will get the other divisor for it.($\because$ if $a$ is a divisor of $n$ then there also exist a divisor $b=\frac{n}{a}$)

Thus each $\frac{n}{x_i}$ term is equal to some other $x_j$ term.

$\frac{1}{n}{\sum }_{i=1}^n{x}_i$

Given, ${\sum }_{i=1}^n{x}_i=75$

Putting that in above, we get

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

$=\frac{75}{n}$

Hence, option B is correct.