There are 9 positive integers with 1 digit (1 - 9).
There are 90 integers with 2 digits (10 - 99).
There are 900 integers with 3 digits (100 - 999).
There are 9000 integers with 4 digits (1000 - 9999).
And so on. It stands to reason that there are [tex]9\times10^{n-1}[/tex] integers with [tex]n[/tex] digits.
Now,
1000 = 9 + 900 + 91
so the 1000th positive integer with an odd number of digits is the 91th number with 3 digits, which is 190.
