How many positive integers are there that have exactly 100 digits in their usual decimal representation?

Introduction / Context:
This question concerns counting how many positive integers have a fixed number of digits, in this case exactly 100 digits. It is a common combinatorial counting problem that tests understanding of place value and the structure of decimal numbers.

Given Data / Assumptions:
- We are asked about 100 digit positive integers in base 10.
- A 100 digit number has its leftmost digit non zero and has exactly 100 places.
- Leading zeros are not allowed because they would reduce the number of digits.

Concept / Approach:
The smallest 100 digit number is obtained by placing 1 followed by 99 zeros, which is 10^99. The largest 100 digit number is the number with 100 nines, which is 10^100 minus 1. The total number of integers from 10^99 to 10^100 minus 1 inclusive gives the count of 100 digit numbers.

Step-by-Step Solution:
Step 1: Identify the smallest 100 digit positive integer. It is 10^99.Step 2: Identify the largest 100 digit positive integer. It is 10^100 - 1.Step 3: The count of integers from A to B inclusive is B - A + 1.Step 4: Here A = 10^99 and B = 10^100 - 1.Step 5: Compute the count: (10^100 - 1) - 10^99 + 1 = 10^100 - 10^99.Step 6: Factor out 10^99 to get 10^99 * (10 - 1) = 10^99 * 9 = 9 × 10^99.Step 7: Thus there are exactly 9 × 10^99 many 100 digit positive integers.

Verification / Alternative check:
We can think digit by digit. For a 100 digit number, the first digit can be any digit from 1 to 9, so there are 9 possibilities. Each of the remaining 99 positions can be any digit from 0 to 9, giving 10 choices each. Therefore the total number of such numbers is 9 * 10^99, which matches the earlier calculation.

Why Other Options Are Wrong:
Option 9 × 10^100 counts numbers with one extra power of 10 and is far too large. Option 10^100 counts all non negative integers with up to 100 digits including shorter ones and zero. Option 11 × 10^98 does not match any correct counting logic for this situation. Option 10^99 counts only the possible combinations when the first digit has a single choice, which is not correct here because the first digit has 9 choices.

Common Pitfalls:
Some learners confuse numbers with at most 100 digits and numbers with exactly 100 digits. Others mistakenly allow the first digit to be zero, which actually reduces the number of digits. Another typical error is to forget that each of the remaining 99 positions has 10 possibilities and not 9.

Final Answer:
The number of positive integers that have exactly 100 digits is 9 × 10^99.