### Instruction: How many 5-digit numbers have at least one 2 or one 3 among their digits?### Response: This problem is about counting the numbers that satisfy a certain condition.
One way to approach this is to use the complementary principle: instead of counting the numbers that have at least one 2 or one 3, we can count the numbers that have neither 2 nor 3 and then subtract that from the total number of 5-digit numbers.
The total number of 5-digit numbers is 90000, since they range from 10000 to 99999.
To count the numbers that have neither 2 nor 3, we can think of how we can choose each digit from the remaining 8 digits: 0, 1, 4, 5, 6, 7, 8, 9.
For the first digit, we have 7 choices, since we cannot choose 0.
For the second digit, we have 8 choices, since we can choose any of the 8 digits.
Similarly, for the third, fourth, and fifth digits, we have 8 choices each.
So the total number of numbers that have neither 2 nor 3 is 7 times 8 times 8 times 8 times 8, which is 28672.