How many numbers are there between 99 and 1000 that have at least one of their digits as 7?
1)648 2)724 3)252 4)814

• Total number of 3 digit numbers having atleast one of their digits as 7 ,
  = (Total numbers of three digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all).
  = (9 × 10 × 10) – (8 × 9 × 9)
  = 900 – 648 = 252.
• 4 years agoHelpfull: Yes(4) No(0)
• between 99 and 1000 there are 901 digits in between
  at least one 7
  0-9 = 9 digits
  _8 * _9 * _9 = 6 or _9 * _8 * _9 or _9 * _9 * _8 = 648
  901-648=253
• 4 years agoHelpfull: Yes(1) No(1)
• let number be ABC
  case1:
  if A=7 then B,C can be (0,1,2,3,4,5,6,8,9)=(1)(9)(9)=81
  
  case2:
  if B=7 then A cannot be 0 because if it is 0 it becomes 2 digit number and C can be any number (0-9 leaving 7)=(8)(1)(9)=72
  
  case3:
  if c=7 then A cannot be 0 and B can be any number leaving 7 =(8)(9)(1)=72
  
  so sum of all case=81+72+72=225
• 3 years agoHelpfull: Yes(0) No(0)
• answer is 252
• 2 years agoHelpfull: Yes(0) No(0)
• Let a 3-digit number ABC
  
  Case (1)
  
  If A=7, then B and C can be any digit but not 7
  
  so, total 9 digits (0,1,2,3,4,5,6,8,9) can replace B and C
  
  Then total such combinations =1(9)(9)=81
  
  Case (2)
  
  If B=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.
  
  And C can be any digit but not 7
  
  Then total combinations =(8)(1)(9)=72
  
  Case (3)
  
  If C=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.
  
  And B can be any digit but not 7
  
  Then total combinations =(8)(9)(1)=72
  
  So, total numbers between 99 and 1000 which have exactly one of their digits as 7 are =81+72+72=225
  
  And 252 answer .
• 1 year agoHelpfull: Yes(0) No(0)