6. Given that the odd number $n$ is a three-digit number, and the sum of the last digits of all its factors (including 1 and $n$) is 33. Then $n$ $=$ . $\qquad$
6.729 (or $27^{2}$).

From $n$ being an odd number, we know that each of its factors is also odd, and of course, the unit digit of each factor is odd. Since the sum of the unit digits of all factors is 33, which is an odd number, $n$ must have an odd number of factors. Therefore, $n$ is a perfect square.

Since a prime number has only 2 factors, and its square has only 3 factors, the sum of the unit digits of these factors is less than 27. Therefore, the three-digit number that meets the condition can only be the square of a composite number. There are 4 such numbers:
$$
27^{2}, 25^{2}, 21^{2}, 15^{2} \text {. }
$$

Upon verification, $27^{2}$, which is 729, is the solution.