when 101102103104............150 is divided by 9 what is the remainder?

• 1(01) 1(02) 1(03)................1(50)
  sum of the digit = 1325
  1325/9 remainder is 2.
  (or)
  101 102 103 104..... 150
  sum of the digit = 150*151/2 - 50*51/2;
  = 11325 - 5050
  = 6275/9
  remainder is 2
  
• 9 years agoHelpfull: Yes(8) No(10)
• sum of the n digits = n/2[a+l].
  here n=50 , a=101 and l=150
  so , 50/2 [101+150]
  25 * 251 = 6275.
  and when 6275/9 , rem=2 .
  hence answer 2.
• 6 years agoHelpfull: Yes(5) No(0)
• when 101/9---remainder will be 2.
  when 101102 is divided by 9 remainder will be 4..
  hence it will increment by 2 every time.
  
  150/3=50.
  that is 50*2=100
• 9 years agoHelpfull: Yes(4) No(12)
• 101 102 103 104..... 150
  sum of the digit = 150*151/2 - 100*101/2;
  = 11325 - 5050
  = 6275/9
  remainder is 2
• 6 years agoHelpfull: Yes(1) No(0)
• The divisibility rule for 9 is sum of the digits is to be divisible by 9. So
  We calculate separately, sum of the digits in hundreds place, tenths place, and units place.
  Sum of the digits in hundreds place: 1 x 50 = 50
  Sum of the digits in tenths place : 0 x 9 + 1 x 10 + 2 x 10 + 3 x 10 + 4 x 10 + 5 x 1 = 105
  Sum of the digits in units place : (1 + 2 + 3 + ...+ 9) x 5 = 225
  So total = 380
  So remainder = 380 / 9 = 2
• 5 years agoHelpfull: Yes(1) No(0)