How many five-digit numbers are there that contain at least one digit 3 and are multiples of 3?

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【Analysis and Solution】
Counting, Exclusion Method, Multiplication Principle.
First, we consider the opposite of "at least one digit 3" which is "no digit 3"; thus, the ten-thousands place has 8 choices (cannot choose $0, 3$), and the thousands, hundreds, and tens places each have 9 choices (cannot choose 3). We know that if a number is a multiple of 3, then the sum of its digits is also a multiple of 3; to ensure the sum is a multiple of 3:
(1) If the sum of the first four digits $\div 3$ has a remainder of 0, then the units place can be $0, 6, 9$, with 3 choices;
(2) If the sum of the first four digits $\div 3$ has a remainder of 1, then the units place can be $2, 5, 8$, with 3 choices;
(3) If the sum of the first four digits $\div 3$ has a remainder of 2, then the units place can be $1, 4, 7$, with 3 choices; that is, regardless of what the first four digits are, the units place always has 3 choices;
Thus, the number of five-digit numbers that do not contain the digit 3 and are multiples of 3 is $8 \times 9 \times 9 \times 9 \times 3=17496$; among the numbers from 10000 to 99999, every 3 consecutive numbers include one multiple of 3; hence, the number of five-digit multiples of 3 is $90000 \div 3=30000$;
The number of five-digit numbers that contain at least one digit 3 and are multiples of 3 is $30000-17496=12504$.