If [log101] + [log102] + [log103] + [log104] + ...... + [log10n] = n, where [x] denotes the greatest integer less than or equal to x, then
(1) 96 ≤ n < 104 (2) 104 ≤ n < 107 (3) 107 ≤ n < 111 (4) 111 ≤ n < 116

• are u sure these are tcs latest ques...
  coz these are mostly CAT ques..
• 11 years agoHelpfull: Yes(2) No(7)
• @moumita:
  the questions given by tcs are mostly from CAT and GMAT only..
• 11 years agoHelpfull: Yes(2) No(7)
• if this is the cat question then this has to be base 10 and not 3 digit numbers ending in n. Ok so this is my solution.
  [log 10 1]= 0; [log (10 10)]=1; [log(10 20)]= 1; [log( 10 30)]=1...; [log(10 100)]=2
  so for 1- 100 summation [log 10 n]= 92 ( you can see that manually)
  after which each [log 10 n] =2; thus making the answer lie between 107 and 108.
  *rest my case*
• 9 years agoHelpfull: Yes(0) No(1)
• From [log101] to [log109], everything is zero.
  From [log1010] to [log1099], all of them are = 1.
  So, there are 90 1’s. [log10100] = 2.
  Let the number of three digit numbers after 100 be x. Then, 90 + 2 + 2x = 100 + x i.e. 92 + 2x = 100 + x i.e. x = 8.
  For n = 108. As that is covered in the values given in the 3rd option, hence that is the answer.
• 7 years agoHelpfull: Yes(0) No(1)