101102103.............150/9......what is the remainder?

• 1 01 1 02 1 03 1 04 .............1 50
  sum of digits = (1+1+1+........+50 times) + (01+02+.......+ 50) = 50 + 50*51/2 = 50 + 25*51
  
  50 + 25*51 /9 => 5 + 7*6 /9 => 47/9 => rem = 2
• 9 years agoHelpfull: Yes(14) No(0)
• 1 01 1 02 1 03 1 04 .............1 50
  sum of digits = (1+1+1+........+50 times) + (01+02+.......+ 50) = 50 + 50*51/2 = 50 + 25*51
  50+25*51=1325.
  1325/9 remainder will be 2.
• 8 years agoHelpfull: Yes(2) No(0)
• At 100 place 1 occur 50 times = 50
  At 10 place 1,2,3,4 occur 10times each and 5 occur only 1 time = 105
  At unit place 1,2,3,4,5,6,7,8,9 occur 5 times each = 225
  Now 50 + 105 + 225 = 380
  Divided by 9
  Remainder 2
  Ans:- 2
• 8 years agoHelpfull: Yes(2) No(0)
• 2 is correct
• 9 years agoHelpfull: Yes(1) No(0)
• remainder = 2
  sum of all no. = 380
  remainder = 380%9 =2
• 8 years agoHelpfull: Yes(1) No(0)
• 2 is correct
  
• 8 years agoHelpfull: Yes(1) No(0)
• 3
  according
  
• 9 years agoHelpfull: Yes(0) No(1)
• what were the options
• 9 years agoHelpfull: Yes(0) No(0)
• (1*50)=50
  0*9 + 1*10 + 2*10 + 3*10 + 4*10 + 5=105
  5(1+2+3+..+9)=5*45=225
  sum of digits=380
  sum of digits when divided by 9 will give the remainder.
  380%9= 2
  therefore 2 is the answer.
• 8 years agoHelpfull: Yes(0) No(0)
• 1---150 ---> n(n+1)/2==11325
  1--100=> 5050
  now 11325-5050= 6975%9= 2
• 6 years agoHelpfull: Yes(0) No(0)