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There are 500 of 3-digit numbers begin with an odd one (100-199, 300-301, 500-599,700-799, and 900-999). In terms of "how many strings of", may filter or expand the above answer.
199
Considering only positive integers then:We are looking at all the numbers between 100 to 999 (inclusive).Or 9 sets of 100.For each set of 100 (e.g. 100 to 199 or 200 to 299)There will be 0+1+2+3+4+5+6+7+8+9 = 45 numbers where this statement holds true.We know this because for 100 to 109 there are none.For 110 to 119 there is 1 (110)For 120 to 129 there are 2 (120 & 121)and this pattern holds all the way to 190 to 199 where there are 9.This will be true for each set of 100 numbers so the answer is 45 * 9 = 405.
To determine how many times the digit 9 is written when writing numbers from 1 to 10000, we can consider the pattern of its occurrence in each place value. In the units place, the digit 9 appears 1000 times (from 9 to 9999). In the tens place, the digit 9 appears 1000 times (from 90 to 99, 190 to 199, and so on). In the hundreds place, the digit 9 appears 1000 times (from 900 to 999, 1900 to 1999, and so on). Therefore, the digit 9 is written 3000 times in total when writing numbers from 1 to 10000.
There are 899 three digit numbers (100 through 999).Each century (grouping of 100 numbers) has 50 odd numbers. (half of the numbers are odd, half are even)So.. 100-199, 200-299, 300-399, 400-499, 500-599, 600-699, 700-799, 800-899, 900-999We have 9 centuries.Number of odd numbers = 9*50 = 450