Correct Option (C)

The digit 5 appears 300 times when writing integers from 1 to 1000.

To determine the total occurrences of the digit 5, a systematic enumeration across positional values (units, tens, and hundreds) for numbers from 1 to 999 is employed, as the number 1000 does not contain the digit 5.

Occurrences in the Units Place: The digit 5 appears in the units place for numbers such as 5, 15, 25, ..., 995. This pattern yields 10 occurrences per block of 100 numbers (e.g., 5, 15, ..., 95 within 1-100). Across 10 such blocks (1-100, 101-200, ..., 901-1000), the digit 5 appears 10 × 10 = 100 times in the units place.
Occurrences in the Tens Place: The digit 5 appears in the tens place for numbers such as 50-59, 150-159, ..., 950-959. This pattern yields 10 occurrences per block of 100 numbers (e.g., 50, 51, ..., 59 within 1-100). Across 10 such blocks, the digit 5 appears 10 × 10 = 100 times in the tens place.
Occurrences in the Hundreds Place: The digit 5 appears in the hundreds place for all numbers from 500 to 599. This range includes 100 distinct numbers (500, 501, ..., 599). Thus, the digit 5 appears 100 times in the hundreds place.

The total number of times the digit 5 appears from 1 to 999 is the sum of these occurrences:

Total = 100 (units) + 100 (tens) + 100 (hundreds) = 300.

Since the number 1000 does not contain the digit 5, the final count remains 300.

Incorrect Options:

Options A (269), B (271), and D (302) are incorrect as they do not correspond to the accurate enumeration of the digit 5 across all positional values within the specified range.