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

• 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.
• 9 years agoHelpfull: Yes(14) No(0)
• z is divisible by 6 only when the value of n is even
  so ans is d.
• 9 years agoHelpfull: Yes(4) No(0)
• 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.
• 9 years agoHelpfull: Yes(4) No(4)
• 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
  
• 9 years agoHelpfull: Yes(3) No(0)
• Ans should be c.
• 9 years agoHelpfull: Yes(2) No(3)
• 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
• 9 years agoHelpfull: Yes(2) No(0)
• n is always divisible by 3
• 9 years agoHelpfull: Yes(0) No(2)
• d
  suppose, if u take n=4 then n is always divisible by 2..........
• 9 years agoHelpfull: Yes(0) No(1)
• 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............
• 9 years agoHelpfull: Yes(0) No(1)
• 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
• 9 years agoHelpfull: Yes(0) No(0)
• 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.
• 9 years agoHelpfull: Yes(0) No(0)