You must pick at least 1001 integers to be sure that at least two of them have a digit in common.
To solve this problem, we can use the Pigeonhole Principle, which states that if you want to distribute n items into m containers and n > m, then at least one container must contain more than one item.
In this case, the "items" are the integers from 100 to 999 (inclusive), and the "containers" are the possible combinations of digits (0-9) in each place (hundreds, tens, and units).
Since there are 10 possible digits for each place, there are a total of 10 * 10 * 10 = 1000 possible combinations.
Therefore, if you pick 1001 integers from 100 to 999, you are guaranteed to have at least one duplicate combination of digits (since you have more integers than there are possible combinations).
This means that at least two of the integers will have a digit in common.
So, you must pick at least 1001 integers to be sure that at least two of them have a digit in common.
Question
How many integers from 100 through 999 must you pick in order to be sure that at least two of them have a digit in common?