Find the 27393rd term of series - 1234567891011121314……..?

Concept: To find the nth digit of a series formed by writing consecutive positive integers one after another, count how many digits each block of numbers (1-digit, 2-digit, 3-digit, ...) contributes, running the cumulative count until it reaches or passes n. The required digit lies inside the number in the block where the cumulative count first covers position n; if the leftover count divides the block's digit-width exactly, the digit is the LAST digit of the number reached by that many full numbers.

Application:

1. Digits from 1-digit numbers (1 to 9): 9 numbers x 1 digit = 9 digits.

2. Digits from 2-digit numbers (10 to 99): 90 numbers x 2 digits = 180 digits.

3. Digits from 3-digit numbers (100 to 999): 900 numbers x 3 digits = 2700 digits.

4. Total digits used up to 999: 9 + 180 + 2700 = 2889 digits.

5. Digits still needed beyond 999: 27393 − 2889 = 24504 digits.

6. Four-digit numbers start at 1000. Dividing 24504 by 4 gives exactly 6126, with remainder 0 — so the 24504th digit is the LAST digit of the 6126th four-digit number.

7. The 6126th four-digit number is 1000 + 6126 − 1 = 7125.

8. The last digit of 7125 is 5.

Cross-check: 6126 complete four-digit numbers occupy exactly 6126 × 4 = 24504 digits, matching the leftover count exactly (zero remainder), which confirms the target digit is precisely the units digit of 7125, i.e. 5.