9^1+9^2+9^3........9^n, where n is the multiple of 11..what will be the remainder when this is divided by 6..
a.2. b.3..c.0 d.none of these..

• d. none of these.
  
  9^1+9^2+9^3........9^n / 6
  
  = 3^1 + 3^2 + 3^3 + .....+ 3^n / 6
  
  = 3 + 3 + 3 + 3 + ......+ 3 / 6 [3^n /6 always gives rem = 3]
  
  = 3n / 6
  
  put n = 11,22,33,44,....
  if n is odd , rem = 3
  if n is even , rem = 0
  
  
• 10 years agoHelpfull: Yes(34) No(0)
• ans is 3
  
  9^11 %6
  = 9^3 %6
  = 3^3%6(9%6=3)
  = 27%6 =3
• 10 years agoHelpfull: Yes(1) No(0)
• ans-3 (9^11)/6 = (3^11)/6 , so remainder will be 3
• 10 years agoHelpfull: Yes(0) No(2)
• ans- none of these.
  even-0 and odd-2
• 10 years agoHelpfull: Yes(0) No(0)
• ans=3
  the units digit for a number in que with n as multiples of 11 is 9
  therefore 9/6=3
• 10 years agoHelpfull: Yes(0) No(0)
• In the question given that n is multiple of 11.
  so upto every 11 term sum of remainder is 33.when we divide 33 with 6 than we get remainder 3.
  so final answer should be 3.....
• 10 years agoHelpfull: Yes(0) No(1)
• i think 9^11 is cyclicity is 4 so remainder 3 then fing 9^3 / 6 which gives remainder 3. with all the multiples of 11.
  
• 10 years agoHelpfull: Yes(0) No(0)