4. How many even five-digit numbers are there that are not divisible by 3 and do not contain the digit 9?
4. The last (fifth) digit must be even, which gives 5 possibilities (0, 2, 4, 6, or 8). For the first digit, there are a total of 8 possibilities (any digit except 0 and 9), and for the second and third digits, there are 9 possibilities each (any digit except 9). It remains to choose the fourth digit. When selecting it, we must ensure that the sum of the digits is not divisible by 3 (so that the resulting number is not divisible by 3). The sum of the digits chosen so far gives a remainder of 0, 1, or 2 when divided by 3; in the first case, we can choose any digit except 0, 3, and 6, which gives 6 possibilities; similarly, in the second and third cases, we have 6 possibilities each (avoiding the digits 2, 5, and 8, or 1, 4, and 7, respectively). Therefore, regardless of the remainder that the sum of the digits chosen so far gives when divided by 3, we always have 6 choices for the fourth digit. Thus, the total number of such numbers is $8 \cdot 9 \cdot 9 \cdot 6 \cdot 5=19440$.