To find the number of 3-digit numbers greater than 500 that have at least one digit equal to 9, we can use the principle of complementary counting:

Total 3-digit numbers greater than 500:

• The smallest 3-digit number greater than 500 is 501.
• The largest 3-digit number is 999.
• The total number of 3-digit numbers greater than 500 is: $999 - 500 + 1 = 500$

Total 3-digit numbers greater than 500 without the digit 9:

To count these, we need to ensure that no digit in the number is 9:

• Hundreds place: The possible digits are 5, 6, 7, or 8 (4 choices, since 9 is excluded).
• Tens place and Units place: The possible digits are 0, 1, 2, 3, 4, 5, 6, 7, 8 (9 choices each).

So, the total number of 3-digit numbers greater than 500 with no 9s is: $4 \times 9 \times 9 = 324$

Total 3-digit numbers greater than 500 with at least one 9:

By complementary counting: $500 - 324 = 176$

Final Answer:

There are 176 three-digit numbers greater than 500 that have at least one digit equal to 9.

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Would you like a breakdown of examples or more specific explanations? Here are some related questions:

1. How can we list all these numbers explicitly?
2. What if the restriction changes to "at least two digits are 9"?
3. How do the calculations change for 4-digit numbers instead?
4. What if the range starts at 600 instead of 500?
5. Can this be solved using permutations and combinations?

Tip: Complementary counting is often easier than directly counting the desired cases.