min num of digits required to form every num greater than 900 and less than 9000.
a.39
b.72
c.49
d.81

• 10+10+10+9=39
• 9 years agoHelpfull: Yes(2) No(9)
• please explain it clearly
  
• 9 years agoHelpfull: Yes(2) No(0)
• 1) for unit place we take 0 to 9 so =>10 digit
  2)for 10nth place same as 0 to 9 so => 10 digit
  3)for 100th place digit ==========>10digit
  4)at last 1000nd digit (9000)====>9
  so,10+10+10+9=39
  
• 9 years agoHelpfull: Yes(2) No(0)
• 900
• 9 years agoHelpfull: Yes(0) No(2)
• it must be (a) i.e., 39.
  10+10+10+9
• 9 years agoHelpfull: Yes(0) No(0)
• 10+10+10+9=39
• 9 years agoHelpfull: Yes(0) No(0)
• plz explain properly
  
• 9 years agoHelpfull: Yes(0) No(0)
• 
  10+10+10+9=39 it is only for 1000 to 9000 na
  what is about 900 to 1000
• 9 years agoHelpfull: Yes(0) No(0)
• case 1 : 900-999
  units place --- 9 digits [(1-9) since>900]
  tens place --- 10 digits [0-9]
  hundreds place - 1 digit [only 9 is possible]
  
  sum=1+9+10=20
  
  case --2 : 1000----9999
  
  units place---10 digits [0-9]
  tens place----10 digits [0-9]
  hundreds place-- 10 digits [0-9]
  thousands place ---8 digits [ 0 is prohobited because it becomes 3 digit number ex : 0345 and 9 is also prohobited since
• 9 years agoHelpfull: Yes(0) No(0)