find remainder of (9^1+9^2+.........+9^n)/6
n is multiple of 11.
a) 0
b) 5
c) 3
d) can not be determine

• Correction in my previous solution.
  If we take the even multiple of 11, then remainder will be zero.
  Therefore answer is cannot be determine.
• 7 years agoHelpfull: Yes(26) No(3)
• 9/6 remainder is 3
  9^2/6 remainder is 3
  9^3/6 remainder is 3
  9 to the power of any number when divided by 6 ,the remainder will always be 3.
  Now, ( 3+3+3 ......11 times)/6 =(3*11)/6; therefore the remainder will be 3.
• 7 years agoHelpfull: Yes(18) No(4)
• Answer will be 3....
  All the 9^i values end with 9 or 1.. All are odd values...
  So (9^1+9^2+.........+9^n) will be multiple of 3 and multiple of 2 (so multiple of 6) if n is even..
  if n is odd sum will be odd , but multiple of 9 , simply multiple of 3....
  So , it will give remainder 3....
• 7 years agoHelpfull: Yes(15) No(3)
• ans is cannot be determined
  9/6 always remainder 3
  (9^2)/6 remainder 3
  therefore if n = 11 --- sum = (11*3)/6, remainder = 3
  if n = 22 sum = (22*3)/6, remainder = 0
  
• 7 years agoHelpfull: Yes(10) No(0)
• cannot be determined as we have both options 0 and 3 possible as remainder
• 7 years agoHelpfull: Yes(6) No(0)
• rem[(9^1 + 9^2 + 9^3 +....9^n)/6]=rem[(3^1 + 3^2 + 3^3 +....3^n)/6]
  1) now if n=11
  rem[3*11/6]=3
  2) if n=22
  rem[3*22/6]=0
  
  so we r getting different values of remainder.. So correct option is d
• 7 years agoHelpfull: Yes(4) No(1)
• Answer depend upon n if n is odd then answer will be 3 if n is even answer will be 0
• 7 years agoHelpfull: Yes(3) No(0)
• it can be rewritten as 9^1/6 +9^2/6+9^3/6 ....
  now n can be
  a) even ..say n= 22
  so
  the rule follows 9^1/6=9/6=3/6 i.e the remainder of remainder.
  so 9^22/6= 9x9x9x9...22 times/6 or 3x3x3...22 times/6 or 3^22/6 which is ((3^4)^5 x 3^2)/6 which gives unit digit as zero
  following same terms
  n = odd say 11
  unit digit will be 3
  so answer is 0 or 3 depending on n
• 7 years agoHelpfull: Yes(2) No(0)
• Answer will be 5
  (9^1+9^2+........+9^11)/6=((3)^2+(3)^4+...........+(3)^22)/6=3((3)^1+(3)^3+...........+(3)^21)/3*2=(3)^1+(3)^3+...........+(3)^21)/2=now last digit will be zero thats why reminder 0 after divided by 2.
  
• 7 years agoHelpfull: Yes(1) No(8)
• 9*11=99
  
  99/6=== reminder is 3
• 7 years agoHelpfull: Yes(0) No(9)
• a 0 last digit is always 9 so
• 7 years agoHelpfull: Yes(0) No(2)
• ANS is 3
  9*(1+2+3+11n)/6=9*(11n(11n+1)/2)/6 (suppose it is 11)
  9*11(11+1)/6*2=3(ans)
  9*22(22+1)/6*2=3(ans)
• 7 years agoHelpfull: Yes(0) No(5)
• correct answer is 3
  here n=11
  if we divide 9/6 we get remainder =3
  '' 9^2 /6 =3
  .....
  
  ..
  ...
  .....
  go to last 9^11/6 =3.
  so in last u get remainder=3
  
• 7 years agoHelpfull: Yes(0) No(5)
• 9^1--last digit=9
  9^2--last digit=1
  9^3--last digit=9
  9^4--last digit=1
  9^5--last digit=9
  9^6--last digit=1
  9^7--last digit=9
  9^8--last digit=1
  9^9--last digit=9
  9^10--last digit=1
  9^1--11last digit=9
  ----------------------------
  sum of last digit=9
  9/6 leaves remainder 3..
  so 3 is answer
  9^2--last digit=1
• 7 years agoHelpfull: Yes(0) No(1)
• (91 +92 +...+9n )mod 6 = (31+32+...+3n)mod 6
  
  =(3+9+9*3+.....+3n)mod 6
  
  every term has remainder 3
  
  so n=11
  
  (3*11)mod 6=33 mod 6=3
• 6 years agoHelpfull: Yes(0) No(1)