Suppose you write down all the numbers 1, 2, 3, 4, ..., 9, 10, 11, ... The tenth digit you write is '1' and the eleventh digit is '0', as part of the number 10. (a) What would the 1000th digit be? At which number would it occur? (b) What number would contain the millionth digit? (c) When would you have written the digit '5' for the 5000th time?
The 1000th digit would be '3' occurring in the number 370. Single-digit numbers give 9 digits, two-digit numbers give 180 digits (total 189). We need 811 more digits from three-digit numbers. 811 ÷ 3 = 270 remainder 1, so it's the first digit of the 271st three-digit number, which is 370.
The millionth digit would be '1' in the number 185185. After single-digit (9), two-digit (180), three-digit (2700), four-digit (36000), and five-digit (450000) numbers, we have 488889 total digits. We need 1000000 - 488889 = 511111 more digits from six-digit numbers. 511111 ÷ 6 = 85185 remainder 1. So it's the first digit of the 85186th six-digit number, which is 100000 + 85185 = 185185.
The digit '5' appears in units place every 10 numbers, in tens place every 100 numbers (for 10 consecutive numbers), etc. This requires careful counting of occurrences of '5' in all positions.
Explanation
(a) Systematic counting of digits in the sequence. (b) Continue counting digits through multi-digit numbers. (c) Count occurrences of digit 5 in units, tens, hundreds, etc. positions.