How many integers between 0 and 1000 DO NOT contain the integer 1

• from 1 to 99 = 19 integers having 1
  from 101 to 199 = all are having 1
  similarly 201-299, 301-399, 401-499,.....so on, will have 19 integers having 1
  
  Answer = 1000 - (19*9 + 99 + 2) // 2 is added for 100 and 1000
  = 728
• 12 years agoHelpfull: Yes(17) No(6)
• Base 9 will have 9^3 = 729 numbers from 0-999.
  here integers between 0 and 1000 are required.
  so Zero is excluded.
  rest 729-1=728 numbers will not have digit 1.
• 12 years agoHelpfull: Yes(13) No(4)
• 728 integers between 0 and 1000 do not contain integer 1.
• 12 years agoHelpfull: Yes(3) No(1)
• Total no. of integers between 0 and 1000 = 999
  By per and comb ,
  1 digit number not containing number '1' = 8
  2 digit number not containing number '1' = 72
  3 digit number not containing number '1' = 648
  Therefore ,total number of integers nit containing '1' = 648 + 72+ 8 = 728
• 8 years agoHelpfull: Yes(3) No(0)
• 799. because from 0-100 it is 80, 101 to 999 it is 719. thus 719+80=799
• 11 years agoHelpfull: Yes(1) No(0)
• is there any relevant formula garima??
• 12 years agoHelpfull: Yes(0) No(2)
• 728
  101 to 199 contains 100 numbers having 1 in it.. 0-99, 200-299 and so on each contains 19 numbers having 1 in it.. again 1000 has 1 in it.. so answer is
  {1000-((19*9)+100)}-1=728
• 11 years agoHelpfull: Yes(0) No(0)
• the answer is 271
• 9 years agoHelpfull: Yes(0) No(1)
• 271.. 1 to 99 -19 no
  among 100to 999 suppose no no having 1 so such combination 8*9*9=648
  so having 1 900-648=252
  total 252+19=271
  
• 8 years agoHelpfull: Yes(0) No(0)
• Answer = 709
  
  1 to 99 = 19 integers have 1
  100 has 1
  101 to 199 = 99 has 1
  201-299, 301-399, ....901-999 has 19 each
  100 has 1
  
  total number of integers from 0 to 1000 having 1 = 19 + 1 + 99 + 19*9 + 1 = 291
  Required ans = 1000 - 291 = 709
• 8 years agoHelpfull: Yes(0) No(0)