What will be the rem.-
1011021030104105............150. divided by 9?
option-
0
2
8
4

• ans:2 because sum of the terms must be divisible by 9
  101+102+103+104+105+....150, n/2*(first term +last term)
  50/2*(101+150)
  25*251=6275
  697*9=6273
  rem.=6275-6273=2
  
• 9 years agoHelpfull: Yes(16) No(5)
• if a number is divisible by 9 then sum of its digits should be divisible by 9.
  SO now from 101 to 150 the first digit is 1 so the sum is 1+1+....1=50,
  And then taking two digits at a time sum of first 50 natural numbers is n(n+1)/2 which gives
  50(50+1)/2= 1275 , 1275+50=1325, now 1325%9=2.
  
  if we sum up single digits it gives 380, 380%9=2
  so the remainder is 2
• 9 years agoHelpfull: Yes(1) No(0)
• ans is 0
  if we add all the digit then divide the sum by 9 then it gives remainder 0
  
• 9 years agoHelpfull: Yes(0) No(1)
• PAPOLU How you got this value 697*9=6273
• 9 years agoHelpfull: Yes(0) No(0)
• 101+102+103+104+105+....150, n/2*(first term +last term)
  50/2*(101+150)
  25*251=6275
  6275/9= gives rem 2 so ans.=2
• 9 years agoHelpfull: Yes(0) No(1)
• 101+102+........+150 mod 9
  = 50(101+150)/2 mod 9
  = 25*251 mod 9
  = (-2)(-1) mod 9 = 2
• 9 years agoHelpfull: Yes(0) No(1)
• 100th digit: 1*50=50
  10th digit:1*10+2*10+3*10+4*10+1*5=105
  1st digit:(1+2+3+4+5+6+7+8+9)*5=225
  Now add all ditits sum=290
  Now add 2+9+0=11
  11/9=2 Answer
  
• 9 years agoHelpfull: Yes(0) No(1)
• ANS : 2
  By basic method : {using sum of n number formula }
  
  = (1+2+3+---+150) - (1+2+3+...+100)-.................................................................we get {101+102+.....+150}
  = (150*151)/2 - (100*101) /2
  = 11325 - 5050
  = 6275
  
  6275 divided by 9 and get remainder as 2
  
  > {2}
• 9 years agoHelpfull: Yes(0) No(0)