When all the integers from 1 to 1000 (inclusive) are written down, how many times does the digit 8 appear in total?

Introduction / Context:
Digit counting problems are common in aptitude tests and help develop careful combinatorial reasoning. Here we want to know how many times the digit 8 appears when every integer from 1 to 1000 is written. Each integer can contribute zero, one or more occurrences of the digit 8, and we must count all of them across the entire range.

Given Data / Assumptions:

• We consider all integers from 1 to 1000 inclusive.
• The numbers are written in standard decimal notation.
• We count each occurrence of the digit 8 in any position (units, tens, hundreds).
• The number 1000 contains no digit 8.

Concept / Approach:
It is easier to treat numbers from 000 to 999 and count occurrences of 8 in each digit place, then adjust for the inclusion of 1000. For 000 to 999, there are 1000 numbers with three positions (hundreds, tens, units). In each position, each digit from 0 to 9 appears equally often. Once we know how many eights appear in each position, we sum across positions. Finally we note that 1000 does not add any eights, so the count remains unchanged for the range 1 to 1000.

Step-by-Step Solution:
Step 1: Consider all 3 digit strings from 000 to 999. There are 1000 such numbers, and each position (hundreds, tens, units) cycles through the digits 0 to 9 equally. Step 2: Count the number of times digit 8 appears in a fixed position. In any fixed position, each digit occurs in exactly 1 out of every 10 numbers. Therefore, in 1000 numbers, 8 appears 100 times in each position. Step 3: There are 3 positions (hundreds, tens, units). Total occurrences of 8 from 000 to 999 = 100 (hundreds) + 100 (tens) + 100 (units) = 300. Step 4: Adjust from 000 to 999 to 1 to 1000. The leading 0 in 000 does not change the count of digit 8 compared with starting at 1. The number 1000 has no digit 8, so the total remains 300.

Verification / Alternative check:
To confirm, we could test a smaller range, such as 1 to 100, and verify that the same positional argument works there. For 1 to 100, each digit appears 20 times in each position among the numbers 00 to 99, leading to 20 eights per position and 60 in total. Extending this reasoning to 000 to 999 is logically consistent. The absence of any 8 in 1000 makes the 300 count final.

Why Other Options Are Wrong:
The options 100 and 200 underestimate the count because they consider only one or two positions or miscount the frequency of 8. The value 400 would overcount, as it implies more than 100 eights per position in 1000 numbers, which is impossible. The option 180 is similarly too small and does not align with equally distributed digits.

Common Pitfalls:
A typical mistake is to attempt to manually count ranges like 80 to 89 or 800 to 899 and double count overlaps such as 888. Others might forget that numbers less than 100 or 10 can be treated with leading zeros to apply the positional method. Another common error is to ignore that each digit appears equally often in each place when all numbers are considered systematically. Using the positional counting method avoids such errors.

Final Answer:
The digit 8 appears a total of 300 times when the integers from 1 to 1000 are listed.