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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 `underset(i=1)overset(k)(sum)(1)/(x_(i))` is equal to

A

`(75)/(k)`

B

`(75)/(n)`

C

`1/n`

D

`(1)/(75)`

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The correct Answer is:
B
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