The string of digits 135791113...999 is formed by

single digit odd will be 5
2 digit odd will be 11 to 99 ; 2*5*9 ; 90
95 digit string from 1 to 99
110-95 ; 15
101 103 105 107 109 ; 9 will be the 105th digit from left
option E

Talking about positive integers, first we have 9 one digit integers (from 1 to 9), then 90 two digit integers (from 10 to 99) and then 900 three digit integers.

If you are not sure how to calculate, say how many two digit integers we have, you know that we have 10 to 99 two digit integers. Number of integers in this case is 99 - 10 + 1 = 90
(We take + 1 because 10 is also a two digit integer)

Of the 9 one digit integers, 5 are odd because we start with 1 and end with 9 (both odd).
Of the 90 two digit integers, 45 are odd because we start with 10 and end with 99 (one even one odd so equal split)

This accounts for total 5 + 2 * 45 = 95 digits.

We still have another 15 digits to go. Now all numbers will be three digit numbers. I simply need the next 5 three digit numbers.
101, 103, 105, 107, 109

So the 110th digit is 9.

Answer (E)

From 1 to 99 there are 50 odd integers, consisting of 5 single-digit and 45 two-digit odd integers.

So far, that’s a total of 5 + 45 × 2 =95 digits.

We still need 110 – 95 = 15 digits, which can be exactly provided by the first five (=15/3) three-digit odd integers.

The fifth three-digit odd integer is 109, whose last digit is 9.

Therefore, the 110th digit in question is 9.

Answer: E

Can anyone explain what the meaning of "decimal representations" in this question? Seems these numbers are represented as integers being combined to form a continuous string.

An alternate approach is to write out the digits in ROWS OF 10 and look for a pattern:

1 3 5 7 9 1 1 1 3 1
5 1 7 1 9 2 1 2 3 2
5 2 7 2 9 3 1 3 3 3

As illustrated by the digits in blue, the row endings exhibit the following pattern:
1st row --> 31 (where 1 is the tens digit for 15)
2nd row --> 32 (where 2 is the tens digit for 25)
3rd row --> 33 (where the second 3 is the tens digit for 35)
Implication:
9th row --> 39 (where 9 is the tens digit for 95)

The 9th 10-digit row brings the total number of digits to 90.
New two rows, beginning with the units digit for 95:
5 9 7 9 9 1 0 1 1 0
3 1 0 5 1 0 7 1 0 9

The digit in green is the 110th in the string.

Base-10 notation, also known as decimal notation, is a method of representing numbers using ten digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. This is the conventional system for denoting numbers, in contrast to systems like the binary numeral system, which is a base-2 number system.

So, the odd integers from 1 through 999, written as the string of digits would simply be odd numbers written one after another: 135791113...999

I attached my work to this problem. Hope it is helpful!

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