What will be the remainder of 10102103104105.........150 when divided by 9 ?

• using formula of AP ,
  Sum = n/2 (a+l) [here n=number of element , a=1st term,l=last term ]
  = 50/2 * (101+150)
  =6275
  
  6275 mod 9 = 2
• 9 years agoHelpfull: Yes(14) No(4)
• Here simple way for division by 9 we need sum,of all dgts thus it is required to count values
  In hundred place 1 will be 50 times so 50
  0 in tens place 10 time 0*10=0
  Looking for 1 to 9 in unit place (100-109,110-119,120-129,130-139,140-149)sum of all values from 1-9 nd 1-9 vlues are repeating 5 time so
  =45*5=225
  Again,looking for 1,2,3,4, 5 in tens place 110-119,120-129,130-139,150-149,150 1*10+2*10+3*10+4*10+5*1 Now,adding all values
  225+50+105 =380
  Now 380/9=2 remainder
  
• 9 years agoHelpfull: Yes(6) No(2)
• sry for d typing mistake
  What will be the remainder of 101102103104105.........150 when divided by 9 ?
• 9 years agoHelpfull: Yes(2) No(1)
• Is it 101+102+103......+150/9 ??
  
  If it is not so then how can you use AP formula for this??
• 9 years agoHelpfull: Yes(2) No(3)
• the remainder is 3
  
• 9 years agoHelpfull: Yes(1) No(8)
• Its almost like an ap series, you can check by any value.
  
  @Rich , You are right . I also thought about that.
• 9 years agoHelpfull: Yes(1) No(0)
• the nos are arranged as 10102103104.......150. According to the AP series we can start from 102 not 101 as the series is 10102 , we can consider it as 1 in the 100's place *50 = 50.
  next,0 in the 10th place is 8 times and den 1 is for 10 times =10. Therefore 10+(2*10)+(3*10)+(4*10)+(5*10)=150 again when we consider the unit's place we get (0+1 2+3+4+5+6+7+8+9)=45. As it repeats from 102 to 150 i.e for five times i.e 45*5=225-1= 224, as the series is from 102 and not 100.
  Now adding all 50+150+224=424
  now 424%9= 1. I might be wrong but logically i think this should be the ans.
• 9 years agoHelpfull: Yes(0) No(0)
• 6. If 3
• 9 years agoHelpfull: Yes(0) No(2)