123412344123444123444412344444... now the last two terms of 200th number

• Here in this question 1234123441234441234444....
  Here after 123 the no of 4s varies arithmetically so no of 4s varies directly to the no of 123s before it.
  Suppose i take 17 4s at end .
  So total 4s in it is 17*18/2=153.
  Now 123 pairs = 17 so total nos =17*3=51+153=204
  Now subtract 4 from these . Still 13 4s .
  So last 2 digits are 44.
  Hope it is the right one .
  come with another one if any.
  
• 8 years agoHelpfull: Yes(18) No(2)
• 441....I THINK SO
• 8 years agoHelpfull: Yes(4) No(4)
• 1st 4 terms 1s - 1, 2s - 1, 3s - 1 4s - 1 = 4 = 4*1+0
  1st 9 terms 1s - 2, 2s - 2, 3s - 2 4s - 3 = 9 = 4*2+1
  1st 15 terms 1s - 3, 2s - 3, 3s - 3, 4s - 6 =15= 4*3+3
  .
  .
  .
  1st 188 terms 1s - 18, 2s - 18, 3s - 18 4s - 134 = 188 = 18*4 + 116
  
  so till 188 terms its 4 then again 1234444444......
  so last two terms of 200th number is 44
• 8 years agoHelpfull: Yes(3) No(1)
• 1st group 1234 number of digit 4
  2nd group 12344 number of digit 5
  3rd group 1234444 number of digit 6
  so at the end of each group there will be series of 4s.
  So the answer is 44
  
• 8 years agoHelpfull: Yes(3) No(2)