3^0 + 3^1 + 3^2 + . . . . .201 terms is divided by 13

a. 0
b. 12
c. 3
d. None of these

• ANS is 0
  Lets see this cycle
  3^0%13=1
  3^1%13=3
  3^2%13=9
  3^3%13=1
  3^4%13=3
  3^5%13=9
  ..... and so on, as it repeat after 3 power, so remainder remain same for more powers
  Now for adding 1st three remainder =1+3+9=13/13=0
  after every three elements reaminder is 0
  201/3=0 thus overall remainder is 0
  
• 9 years agoHelpfull: Yes(53) No(3)
• Ans is 0
  
  3^0+3^1+...210 terms is geometric progression with 201 terms the sum of which is calculated using formulae to be 3^201 - 1/2
  
  Now 3^201 is nothing but 3^20 X3^20..ten times X 3 and 3^20 has unit digit 1 so obviously the unit digit of while expression is 3. Thus 3^210 - 1 has unit digit 2 which when divided by 2 will give no remainder. Hence remainder is 0.
• 9 years agoHelpfull: Yes(5) No(2)
• series =(1+3+(3^2)+(3^3)+(3^4)+(3^5)......)/13
  ({1+3+(3^2)}+(3^3){1+(3^1)+(3^2)}......)/13
  {13+(3^3)13+......}/13
  {1+3^3+.....}
  thus no remainder
  
• 9 years agoHelpfull: Yes(3) No(0)
• the answer is zero
  3^0=1 and 1%13=1
  3^1=3 and 3%13=3
  3^2=9 and 9%13=9
  and 1+3+9=13 and 13%13=0
  so the result is zero
• 9 years agoHelpfull: Yes(1) No(0)
• 
  last term :
  3^201 mod 13 => 201 mod 12 =9
  3^9 mod 13 =1
  
  sum=(3^201-1)/2=1-1/2=0
  
• 9 years agoHelpfull: Yes(0) No(1)