9^1+9^2+9^3+.........+9^n when divide by 6 what will be the reminder. n should be multiple of 11 ?

1)0
2)1
3)3
4) Can't determine

• 9^1 mod 6 = 3
  9^2 mod 6 = 3
  9^3 mod 6 = 3
  So for all the powers of 9, the remainder is same . i.e. 3 only....
  So 3+3+3+3... 3 mod 6
  If n=11, 11(3) = 33 mod 6 = 3
  If n=22, 22(3) = 66 mod 6 = 0
  If n=33, 33(3) = 99 mod 6 = 3
  So for all odd multiples of 11, i.e. 1*11, 3*11, 5*11, etc, the remainder ll be 3...
  For all even multiples of 11, the remainder ll be 0...
  Hence we can't determine without knowing the exact value of "n"...
  
  Ans : 4) Can't determine
  
• 10 years agoHelpfull: Yes(87) No(5)
• 9^1+9^2+9^3+.........+9^n/6
  3^1+3^2+3^3+...........+3^n/6 (According to remainder theory) a/b= (a-b)/b
  take two term
  (3^1+3^2)+(3^3+3^4)+..........+(3) /6 ( because n multilpe of 11 so if u take two term 1 is left here)
  12/6 +78/6 +............+3/6
  0+0+0+.............+3
  =3
  Ans should be 3
  
• 10 years agoHelpfull: Yes(12) No(18)
• In any case,
  when n is add.. remainder is 3.
  when n is even.. remainder is 0.
• 10 years agoHelpfull: Yes(10) No(0)
• ANSWER WILL BE 3
  BECAUSE
  on dividing 9/6=3(remainder)
  final series will be
  9^1+9^2+9^3..........9^n/6
  will givw
  =3^1+3^2+3^3...3^n
  let n=11
  =3+9+27+81+.......3^11/6
  on adding even terms and dividing by 6 we are left with final odd term 3^11 now
  =3^11/6=(9^5).3/6
  =(3^5).3/6
  =243.3/6
  =3.3/6
  =9/6
  =3 ans
  
• 10 years agoHelpfull: Yes(2) No(2)
• (9^1+9^2)/6 the remender is 0.(9^3+9^4)/6 remender is 0.if we take n=11 thn remender is 3 bt if we take n=22 remender is 0.so the anwer shd be d) Can't determine.
• 10 years agoHelpfull: Yes(2) No(0)
• @Saraswathy
  your's answer is correct but u did mistaken in 2nd line that is......
  9^2 mod 6=0 plz correct it.....
• 10 years agoHelpfull: Yes(1) No(3)
• 2nd option is 2
  
• 10 years agoHelpfull: Yes(0) No(4)
• question was
  9^1+9^2+.....9^n is divisible by 6 nd 6 is multiple of 11
  answer is first option
  1)o.
• 10 years agoHelpfull: Yes(0) No(6)
• question was
  9^1+9^2+.....9^n is divisible by 6 nd "n" is multiple of 11
  answer is first option
  1)o.
  they had asked for the value of n , not for the reminder
  
• 10 years agoHelpfull: Yes(0) No(3)
• ANS:1)0because 9^1+9^2+9^3..........+9^n/6=3^1+3^2+3^3+3^4+..........3^n/6
  where n is a multiple of 11.let it be 11 for demonstration...
  now,on solving powers:
  =3+9+27+81..........3^11/6
  on adding odd terms starting from the first and 3we will get apattern as:
  =30+90+............3^11/6=0
• 10 years agoHelpfull: Yes(0) No(3)
• what does 2^(n-1)gives?
• 10 years agoHelpfull: Yes(0) No(0)
• As we know that last digit of odd power of 9 is 9 and last digit of even power of 9 is 1. from 1 to 11 FIVE(2,4,6,8,10) is even number i.e. last digit (1,1,1,1,1).
  and SIX(1,3,5,7,9,11) is odd power i.e. last digit(9,9,9,9,9,9). mean that last two digit of this sum 9^1+9^2+9^3+.........+9^n will be (54+5)=59 and last digit
  will be 9 . when we will divide this sum by 6 ,reminder will be get 3 (9/6)
• 10 years agoHelpfull: Yes(0) No(0)
• n=11k;
  so if 11k is evn nuber then reainder =0;
  because 3*22%6=0
  3*44%6=0
  
  but if n=11k is odd .eg:11,33,55 etc then
  3*11%6=3
  
  so ans will be (d)
  
• 9 years agoHelpfull: Yes(0) No(0)