Latest Elitmus Aptitude Question SOLUTION: 9^n+9^n-1+........+9^3+9^2+9^1. if n is greater than and equal to 4 (i.e 4,5,6,....) then for all values of n series is always divisible by a. 7380 b. 729 c. 90 d. 9

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(#M40160091) ELITMUS Numerical Ability question Number System Keep an EYE

9^n+9^n-1+........+9^3+9^2+9^1. if n is greater than and equal to 4 (i.e 4,5,6,....) then for all values of n series is always divisible by

a. 7380
b. 729
c. 90
d. 9

Asked In Elitmus pankaj yadav (7 years ago)

Unsolved Read Solution (5)

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Read Solution (5) :

Ekant Chawla  (7 years ago)

as it is a GP, with a =9^n and r=1/9 the sum would be , (9^n(1-(1/9)^n))/(1-1/9). on solving we get ,

9^n/(8/9) = 9^(n+1)/8

now when you put n=4 you will notice that the result will always be divisible by 9^5 and lower but as the options have 9^3=729 and 9^1=9 you could go with either of them.

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DEEPALI AGARWAL  (7 years ago)

taking n=4,
9^4+9^3+9^2+9^1=9(9^3+9^2+9^1+1) , which is divisible by 9

taking n=5 we have 9^5+9^4+9^3+9^2+9^1=9(9^4+9^3+9^2+9^1+1)
which is again divisible by 9

so, the answer is d.

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Amit  (7 years ago)

taking in reverse
9+9^2+9^3+9^4
which is a gp
so sum is 9(9^3-1)/(9-1)=91*9 which is only divisible by 9.

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pankaj yadav  (6 years ago)

Yes, 9 is the right answer.

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Prateek Gupta  (4 years ago)

d
let n=4
9(9^3+9^2+9^1+9^0) is always divisible by 9

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