Question : Given sequence 91011....33 .what is the remainder when divided by 9?
a)2
b)0
c)3
d)7

• c)3
  (9+10+11+12.......33)-(1+2+3...8)
  so here
  n(n+1)/2-n(n+1)/2
  (33x34)/2-(8x9)/2
  33x17-4x9
  561-36=525
  now 525/9=3
  
  
• 8 years agoHelpfull: Yes(8) No(2)
• It will be,
  10's place=> 1*10 + 2*10 + 3*4 = 42
  Unit's place=> 9+2(0+1+2+3+4+5+6+7+8+9)+(1+2+3) = 105
  
  So,105+42=147
  147/9=rem(3)
• 8 years agoHelpfull: Yes(5) No(1)
• S@ch!n
  first term = 9
  Last term = 33
  total number of term = 25
  sum = n(a+L)/2
  sum = 25(9+33)/2 = 42*25/2 = 525
  remainder = 525 / 9 = 3
  Ans = 3
• 8 years agoHelpfull: Yes(5) No(1)
• sum of total no of 1 at 10's place =10*1=10
  sum of total no of 2 at 10's place =10*2=20
  sum of total no of 3 at 10's place =3*4=12
  sum of numbers at unit place=45*2=90
  total sum =90+42+9+6=147
  147/9 rem=3
  
• 8 years agoHelpfull: Yes(4) No(0)
• Add 9+10+11+....+33. You'll get 525. Divide it by 9, the remainder is 3.
• 8 years agoHelpfull: Yes(3) No(3)
• Ans :- 0
  
  sum of unit digit => 9 + unit digit of (10 to 30) + unit digit of (31,31,33 )
  = 9 + 2 *(1+2+3+......+9)+1+2+3
  = 9+ 90+6
  =105
  
  sum of ten's place => ten's place from (10 to 33)
  = 10*1 (from 10 to 19)+ 10*2 (from 20 to 29) +4*3 (for 30,31,32,33)
  =10+20+12
  =42
  total sum of unit and ten's digit = 105+42= 147
  divided by 9 => 147 %9 = 3 will be remainder
  
  
  
• 8 years agoHelpfull: Yes(2) No(3)
• sequence will be
  9101112131415161718192021222324252627282930313233
  divisibility rule of 9 is sum of all digits should be multiple of 9
  sum of digits = 135%9==0;
• 8 years agoHelpfull: Yes(1) No(6)
• 33=a+(n-1)d
  33=9+n-1
  33-8=n
  n=25
  sum=25(9+33)/2
  25*42/2
  25*12=525
  525/9=3
• 8 years agoHelpfull: Yes(1) No(0)
• Answer b) 0
  sum of unit digit= (1+2+....9)*2 +1+2+3+9
  =>sum=45*2+15
  sum=105
  sum of ten's place=10*1 +10*2+ 3*3
  10+20+9
  =39
  => total sum=39+105=144
  which is completely divisible by 9
• 8 years agoHelpfull: Yes(0) No(7)
• Arithmetic progression = 9+10+11+12.....33
  
  where a=9, l=33,
  
  sum of arithmetic progression = n/2(a+l) => 24/2(9+33) =>396
  
  so, 396/9 gives the zero reminder.... Ans= 0
  
• 8 years agoHelpfull: Yes(0) No(6)
• Here, first term,a =9 and last term(l) is 33, n= 25(from 9 to 33).
  Now, Sn= n/2[a+l]
  Sn=23/2[9+33]
  =525
  Now, divide by 9, we get
  525/9= 3(as Remainder).
• 8 years agoHelpfull: Yes(0) No(1)