Latest Elitmus Aptitude Question SOLUTION: z=9^1+9^2+9^3+.....+9^n,z is always divisible by 6 and n is always divisible by------ a) n is always divisible by 9 b)n is always divisible by 6 c)n is always divisible by 3 d) n

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(#M40116551) ELITMUS Numerical Ability question Number system Keep an EYE

z=9^1+9^2+9^3+.....+9^n,z is always divisible by 6 and n is always divisible by------

a) n is always divisible by 9
b)n is always divisible by 6
c)n is always divisible by 3
d) n is always divisible by 2

Asked In Elitmus pravallika (9 years ago)

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DHANANJAY KUMAR  (9 years ago)

d) n is always divisible by 2.z=(9+9^2+9^3....+9^n)/6 => (3+3+3+3+3...+3)/6 .For the remaining elements to be divisible by 6 ,we need the value of n as even i.e.(3+3)/6 would give remainder zero. So , n is always divisible by 2.

Is this solution Helpfull? Yes (14) | No   

Abhishek Saurav  (9 years ago)

z is divisible by 6 only when the value of n is even
so ans is d.

Is this solution Helpfull? Yes (4) | No   

shiv  (9 years ago)

9^1=3^2
9^2=3^4
9^3=3^6
. ^ . .^.
. ^ . . ^ .
now coming to question.
z=9^1+9^2+9^3+.........+9^n is divisible by 6.
=3^2+3^4+3^6+..........+3^2n
=3^2(1^2+3^2+3^4+.....3^2n-2)
now
3^2=9
9*(any term )is divisible by 6.
9*1=9(not divisible by 6.)
9*2=18(divisible by 6)
9*3=27(not divisible by 6.)
9*4=36(divisible by 6)
Hence it is clear that the sum of rest term except 9 is even.
9*(even)=any value divisible by 6.
2n-2 is always even
so option 4 is correct.

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Ayush Singhal  (8 years ago)

Z=9^1+9^2+..........+9^n
For divisible by 6
9/6 gives remainder = 3
(9^2)/6 gives remainder = 3
(9^3)/6 gives remainder = 3
....
....
....
(9^n)/6 gives remainder = 3

so we get remainder
3+3+3+3+3..............n times
=3n
3n is divisible by 6 when n is even
So
d) n is divisible by 2

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ASHISH MUKHERJEE  (9 years ago)

Ans should be c.

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DAMARLA BALAJI  (9 years ago)

ans: d 9^1+9^2 is divisible by 6 and same
9^1+9^2+9^3=9^4 is divisible by 6
when n is even then only it is divisible by 6

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k.s.n.kranthi kumar  (9 years ago)

n is always divisible by 3

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pavan  (9 years ago)

d
suppose, if u take n=4 then n is always divisible by 2..........

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pavan  (9 years ago)

d
Suppose,if u take n=4. Then the total is divisible by 4 and n is divisible by only 2.
So n is always an even no............

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rohit bisht  (9 years ago)

z=(9+9^2+9^3....+9^n)
=9^{n(n+1)}/2
=3^{n(n+1)}
if we divide it by 6 it must give us some +ve integer i.e k
3^{n(n+1)}/6=k

n(n+1) should be such that it gives solution in form of multiple of 6 i.e 6,18,24.....

so eq... reduces in form of
n(n+1)=6k
now substitute each from option one by one option b will give you positive k

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Alok Pandey  (8 years ago)

you will notice that this series will only be divisible by 6 when n has an even power like
9^1 + 9^2 = 90 which is divisible by 6
again
9^1+9^2+9^3+9^4 = 7380 which is again divisible by 6
in both the cases power 2 and 4 gave an answer that satisfies the equation.

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