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

The correct answer is 252.

To find the number of integers in the set {100, 101, 102, ..., 999} that have at least one digit repeated, we can use the principle of complementary counting. That is, we calculate the total number of integers in the set and subtract the number of integers that have no digits repeated (i.e., all digits are distinct).

Step 1: Find the total number of integers in the set
The set consists of all 3-digit integers from 100 to 999.
The total number of integers in this range is:
$$999-100+1=900$$

Step 2: Find the number of 3-digit integers with all unique (non-repeated) digits
Let a 3-digit number be represented as d₁d₂d₃, where each digit is chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.
We choose the digits one by one such that they are all distinct:
1. For the hundreds digit (d₁): It cannot be 0 (since it must be a 3-digit number). Thus, there are 9 possible choices (1 through 9).
2. For the tens digit (d₂): It cannot be the same as the hundreds digit, but it can be 0. Since one digit has already been chosen, there are 9 remaining choices from the 10 possible digits.
3. For the units digit (d₃): It cannot be the same as the hundreds digit or the tens digit. Since two distinct digits have already been chosen, there are 8 remaining choices.

Using the multiplication principle, the number of 3-digit integers with all distinct digits is:
$$9\times 9\times 8=648$$

Step 3: Subtract to find the integers with at least one repeated digit
Subtract the number of integers with distinct digits from the total number of 3-digit integers:
$$900-648=252$$

Therefore, there are exactly 252 integers in the set {100, 101, 102, ..., 999} that have at least one digit repeated.