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

CONCEPT: When natural numbers are written one after another to form a continuous digit string, the position of any digit can be located by counting how many digits each block of numbers contributes: single-digit numbers (1 to 9) contribute 9 x 1 = 9 digits, two-digit numbers (10 to 99) contribute 90 x 2 = 180 digits, three-digit numbers (100 to 999) contribute 900 x 3 = 2700 digits, four-digit numbers (1000 to 9999) contribute 9000 x 4 digits, and so on. Subtracting each block's digit count from the target position narrows down which block, and eventually which exact number and digit, holds that position.

1. Digits contributed by 1-digit numbers (1 to 9): 9 x 1 = 9.

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

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

4. Total digits used up through the number 999: 9 + 180 + 2700 = 2889.

5. Digits still needed to reach position 27393: 27393 - 2889 = 24504.

6. Each four-digit number contributes 4 digits, so 24504 / 4 = 6126 complete four-digit numbers, with remainder 0.

7. The four-digit numbers start at 1000, so the 6126th one is 1000 + 6126 - 1 = 7125.

8. Since the remainder is 0, the target position lands exactly on the last digit of this number, 7125.

CROSS-CHECK: Counting forward confirms this - the 999 numbers before contribute 2889 digits, and the four-digit numbers from 1000 through 7125 contribute (7125 - 1000 + 1) x 4 = 6126 x 4 = 24504 digits. Adding these, 2889 + 24504 = 27393, exactly the target position, confirming the last digit of 7125, which is 5, is indeed the 27393rd digit of the series.