The number of times the digit 5, will appear while writing the integers from 1 to 1000 is

The correct option is 300.

To find the total number of times the digit 5 appears when writing all the integers from 1 to 1000, we can analyze the positions of the digits (units, tens, and hundreds places) in the three-digit numbers. Let us represent any number from 0 to 999 as a three-digit sequence: abc, where each of a, b, and c can be any digit from 0 to 9.

Note that including 0 does not change the count of the digit 5, and the number 1000 does not contain the digit 5. Thus, counting the digit 5 in the range 000 to 999 is equivalent to counting it from 1 to 1000.

We analyze the occurrence of the digit 5 in each of the three positions:
1. Units position (c = 5): The number has the form ab5. Since a can be any of the 10 digits (0-9) and b can also be any of the 10 digits (0-9), the number of such integers is:
$10\times 10=100$
2. Tens position (b = 5): The number has the form a5c. Similarly, a can be any of the 10 digits (0-9) and c can be any of the 10 digits (0-9). The number of such integers is:
$10\times 10=100$
3. Hundreds position (a = 5): The number has the form 5bc. Here, b can be any of the 10 digits (0-9) and c can be any of the 10 digits (0-9). The number of such integers is:
$10\times 10=100$

By summing the occurrences of the digit 5 across all three independent positions, we get the total number of times the digit 5 appears:
$100+100+100=300$