Let `x_(1),x_(2),x_(3), . . .,x_(k)` be the divisors of positive integer 'n' (including 1 and x). If `x_(1)+x_(2)+ . . .+x_(k)=75`, then `sum_(r=1)^(k)(1)/(x_(i))` is equal to

Options

A. `(k^(2))/(75)`
B. `(75)/(k)`
C. `(n^(2))/(75)`
D. `(75)/(n)`

Answer

D

Explanation

`underset(i=1)overset(k)(sum)(1)/(x_(i))=(1)/(x_(1))+(1)/(x_(2))+ . . .+(1)/(x_(k))=(underset(i=1)overset(k)(sum)x_(i))/(n)=(75)/(n)`