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

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Analysis:
Elimination method: First, determine that there are $90000 \div 3=30000$ (numbers) that are multiples of 3 among the five-digit numbers; then determine the five-digit numbers that are multiples of 3 and do not contain the digit 3: the ten-thousands place has 8 choices (1 to 9, excluding 0 and 3), the thousands, hundreds, and tens places each have 9 choices (1 to 9, excluding 3). To make the five-digit number a multiple of 3, when the first 4 digits are determined, if the sum of the first 4 digits is divisible by 3 with a remainder of 0, the units place can take $0, 6, 9$; similarly, if the sum of the first 4 digits is divisible by 3 with a remainder of 1, the units place can take $2, 5, 8$; if the sum of the first 4 digits is divisible by 3 with a remainder of 2, the units place can take $1, 4, 7$; so the units place has only 3 choices, thus there are $8 \times 9 \times 9 \times 9 \times 3=17496$ (ways); therefore, the number of five-digit numbers that meet the conditions is: 30000-17496=12504 (ways).