The question asks how many numbers between 99 and 1000 have at least one digit as 7. Let's break down the solution.

Step 1: Range of Numbers

The numbers between 99 and 1000 are three-digit numbers, ranging from 100 to 999. So, we are looking for numbers from 100 to 999 that contain at least one digit as 7.

Step 2: Complementary Counting

It is easier to count how many numbers do not have any digit as 7 and subtract this from the total number of three-digit numbers.

Total Three-Digit Numbers:

• The total number of three-digit numbers is $999 - 100 + 1 = 900$.

Numbers with No Digit 7:

Each digit (hundreds, tens, ones) can be one of the following digits: 0, 1, 2, 3, 4, 5, 6, 8, 9 (9 choices).

• For the hundreds digit, we can have any number from 1 to 9 except 7. So, we have 8 choices (1, 2, 3, 4, 5, 6, 8, 9).
• For the tens digit, we can have 9 choices (0-9, except 7).
• For the ones digit, we can have 9 choices (0-9, except 7).

Thus, the number of three-digit numbers that do not contain 7 is: $8 \times 9 \times 9 = 648$

Step 3: Subtract to Find Numbers with at Least One 7

Now, subtract the numbers that do not contain 7 from the total number of three-digit numbers: $900 - 648 = 252$

Thus, the number of numbers between 100 and 999 that contain at least one digit as 7 is 252.

Final Answer:

The correct answer is A) 252.

Would you like more details on any part of this solution? Here are five related questions for further exploration:

1. How does the method of complementary counting simplify solving such problems?
2. What if the question asked about numbers containing the digit 5 instead of 7?
3. How many numbers between 1000 and 9999 contain the digit 7 at least once?
4. Can this method be extended to count numbers containing two specific digits (e.g., both 7 and 5)?
5. What is the probability that a random three-digit number contains the digit 7?

Tip: When counting numbers with certain conditions, complementary counting often simplifies the process by focusing on what you don't want first.