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

@moumita:
the questions given by tcs are mostly from CAT and GMAT only..
12 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.
8 years agoHelpfull: Yes(0) No(1)

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