Let n1- n2- - nk be all the factors of a positive integer n including 1 and n - If n1-n2-nk-91- then the value of 1-n1-1-n2- 1-nk is

The correct option is C 91n nbsp;1n_1+ 1n_2 + ⋯ + 1n_k =1n (nn_1+ nn_2 + ⋯ + nn_k) Since n_1, n_2, ⋯ , n_k are factors of n, nn_1, nn_2 , ⋯ , nn_k will also ...

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Standard XII

Mathematics

General Tips for Choosing Function

Let n1, n2, ⋯...

Question

Let $n_{1}, n_{2}, \dots, n_{k}$ be all the factors of a positive integer $n$ including $1$ and $n .$ If $n_{1} + n_{2} + \dots + n_{k} = 91,$ then the value of $\frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}}$ is

\frac{k^{2}}{91}

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\frac{91}{k}

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\frac{91}{n}

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cannot be determined from the given information

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Solution

The correct option is C $\frac{91}{n}$
$\frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}}$
$= \frac{1}{n} (\frac{n}{n_{1}} + \frac{n}{n_{2}} + \dots + \frac{n}{n_{k}})$
Since $n_{1}, n_{2}, \dots, n_{k}$ are factors of $n,$
$\frac{n}{n_{1}}, \frac{n}{n_{2}}, \dots, \frac{n}{n_{k}}$ will also be factors of $n .$

So, $\frac{n}{n_{1}} + \frac{n}{n_{2}} + \dots + \frac{n}{n_{k}} = 91$
$∴ \frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}} = \frac{91}{n}$

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Let n1,n2,⋯,nk be all the factors of a positive integer n including 1 and n. If n1+n2+⋯+nk=91, then the value of 1n1+1n2+⋯+1nk is

The correct option is C 91n 1n1+1n2+⋯+1nk =1n(nn1+nn2+⋯+nnk) Since n1,n2,⋯,nk are factors of n, nn1,nn2,⋯,nnk will also be factors of n. So, nn1+nn2+⋯+nnk=91 ∴1n1+1n2+⋯+1nk=91n

Let $n_{1}, n_{2}, \dots, n_{k}$ be all the factors of a positive integer $n$ including $1$ and $n .$ If $n_{1} + n_{2} + \dots + n_{k} = 91,$ then the value of $\frac{1}{n_{1}} + \frac{1}{n_{2}} + \dots + \frac{1}{n_{k}}$ is