what is the remainder when 123412341234.....upto 400 digits is divided by 909?

• 1234%909=325
  12341234%909=325*2
  1234........400 digits=325*100
  ans=(325*100)%909 = 685
• 11 years agoHelpfull: Yes(58) No(2)
• 685 is d correct ans
• 11 years agoHelpfull: Yes(12) No(2)
• @puneet: please explain
  
• 11 years agoHelpfull: Yes(3) No(0)
• ans :-> 325
  
  123412341234 is completely divisible by 909
  400%909 = 4
  1234 % 909 = 325
• 11 years agoHelpfull: Yes(3) No(8)
• 1
  it is done using g.p. series .
  
• 11 years agoHelpfull: Yes(1) No(7)
• D ans z 315
• 11 years agoHelpfull: Yes(1) No(1)
• @ramil:
  first step 1234(10^396+10^392+10^388.......+1)
  now a g.p. is formed inside with a=1 and r=10^4
  solve for the value of no. of terms in g.p n=100
  find sum of g.p , it will come equal to 999999.......(400times9)/9999
  in next step divide by 101 and you will get {1234*(1010101010.......1)}/9*101
  now 1010101...1 divisible by 101 so you will get remainder by dividing 1234 by 9 and that is equal to 1.
• 11 years agoHelpfull: Yes(1) No(1)
• 1234%909=325
  12341234%909=650=325*2
  .
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  1234 ..... 1000 times = (325*100)%909=685 Ans.
• 11 years agoHelpfull: Yes(1) No(0)
• 1234(4 digit),400/4 gives reminder 0 and q-100
  so the reminder 0
• 11 years agoHelpfull: Yes(0) No(2)
• PRE plz provide us the options..
• 11 years agoHelpfull: Yes(0) No(0)
• my be 200 ???
• 11 years agoHelpfull: Yes(0) No(0)
• 1234%909=325
  12341234%909=325*2
  1234........400 digits=325*100
  ans=(325*100)%909 = 685
  
• 9 years agoHelpfull: Yes(0) No(0)