Question

How many integers in the set {100, 101, 102, ..., 999} have at least one digit repeated?

Answer 1

The total number of 3-digit integers is calculated as:  
\(= 999 - 100 + 1 = 900\)

Let \(N\) be the number of 3-digit integers that have no repeated digits.

We are choosing digits from the set {0, 1, 2, ..., 9}, i.e., 10 digits in total. 
However, for a 3-digit number:

• The hundreds place (first digit) cannot be 0. So we have 9 choices (1 through 9).
• Once the first digit is fixed, we have 9 digits left (excluding the digit used in the first place) for the tens place.
• After choosing the first two digits, we are left with 8 digits to choose from for the units place.

 

Therefore, the number of 3-digit integers with all digits different is: 
\(N = 9 \times 9 \times 8 = 648\)

Hence, the number of 3-digit integers with at least one repeated digit is: 
\(= 900 - 648 = 252\)

Final Answer: \(\boxed{252}\)