the minimum number of digits required to form every number which is greater tha 900 and less than 9000

• The question is given less than 9000 so last number is 8999.So 1000ths digits can be 1-8.It is also given that greater than 900 so 1st no is 901.so unit plac range 1-9=9
  Thus min no of digits =9+10+10+8=37
  
• 9 years agoHelpfull: Yes(7) No(1)
• Answer - 39
  
  for unit place numbers can range from 0-9 ie; total 10 digits
  for tens(10th) place number can range from 0-9 ie; again 10 digits (0,1,2,3,4,5,6,7,8,9)
  similarly for hundred(100th) place again 10 digits.
  but for 1000th place it can only range from 1-9 ie; total 9. We cannot include 0 here as the number will then start from 0 means ex-0459 which only means 459.
  
  thus minimum no of digits required are : 10+10+10+9 =39
  
  39 is the right answer (100% sure)
  
  
• 9 years agoHelpfull: Yes(6) No(1)
• 10 digits
• 9 years agoHelpfull: Yes(0) No(4)
• Ans = 10+10+10+9=39
  
• 9 years agoHelpfull: Yes(0) No(1)
• 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(0) No(1)
• The question is given less than 9000 so last number is 8999.So 1000ths digits can be 1-8.
  Thus min no of digits =10+10+10+8=38
• 9 years agoHelpfull: Yes(0) No(0)
• 39 is the correct......
  
• 9 years agoHelpfull: Yes(0) No(0)
• Answer is weird. Isn't it ? Only 10 digits are present in whole mathematics.
• 9 years agoHelpfull: Yes(0) No(0)