Question

Let $n_1,n_2,\cdots ,n_k$ be all the factors of a positive integer $n$ including $1$ and $n.$ If $n_1+n_2+\cdots +n_k=91,$ then the value of $\frac{1}{n_1}+\frac{1}{n_2}+\cdots +\frac{1}{n_k}$ is
(correct answer + 2, wrong answer - 0.50)

• A. $\frac{k^2}{91}$
• B. $\frac{91}{k}$
• C. $\frac{91}{n}$
• D. cannot be determined from the given information

Answer 1

The correct option is C $\frac{91}{n}$

$\frac{1}{n_1}+\frac{1}{n_2}+\cdots +\frac{1}{n_k}$
$=\frac{1}{n}(\frac{n}{n_1}+\frac{n}{n_2}+\cdots +\frac{n}{n_k})$
Since $n_1,n_2,\cdots ,n_k$ are factors of $n,$
$\frac{n}{n_1},\frac{n}{n_2},\cdots ,\frac{n}{n_k}$ will also be factors of $n.$

So, $\frac{n}{n_1}+\frac{n}{n_2}+\cdots +\frac{n}{n_k}=91$
$\therefore \frac{1}{n_1}+\frac{1}{n_2}+\cdots +\frac{1}{n_k}=\frac{91}{n}$