Question

Let d₁, d₂, ….., d_k be all the factors of a positive integer n including 1 and n. Suppose d₁ + d₂ + …….+ d_k = 72. Then the value of $\frac{1}{d_{1}}$+$\frac{1}{d_{2}}$+……. +$\frac{1}{d_{k}}$ is

• A. $\frac{k^{2}}{72}$
• B. $\frac{72}{k}$
• C. $\frac{72}{n}$
• D. Cannot be computed from the given Information

Answer 1

Answer: (C)

$\frac{72}{n}$

Answer 2

Sol. Explanation:

$\frac{1}{d_{1}}$+$\frac{1}{d_{2}}$+$\frac{1}{d_{3}}$+……+$\frac{1}{d_{k}}$= $\left\lbrack \frac{n}{d_{1}} + \frac{n}{d_{2}} + \frac{n}{d_{3}} + ... + \frac{n}{d_{k}} \right\rbrack$

Now, $\frac{n}{d_{1}}$,$\frac{n}{d_{2}}$,…… will also be divisor of the number i.e., $\frac{n}{d_{j}}$= d_l for same j and l

Ž$\frac{1}{d_{1}}$+ $\frac{1}{d_{2}}$ +……..+$\frac{1}{d_{k}}$= $\frac{1}{n}$ [d₁ + d₂ +…….] =$\frac{72}{n}$