7.2. Find the sum of all four-digit natural numbers composed of the digits 3, 6, and 9.
Answer: 539946. Solution. We will add the numbers in a column. Each last digit (for example, six) appears in the units place as many times as there are three-digit numbers of the specified type with this last digit (in our case -- these are numbers of the form $\overline{x y z 6}$, where $x, y, z$ are arbitrary sets of digits 3, 6, and 9. Thus, the digit six, like three and nine, will appear in the units place $3 \cdot 3 \cdot 3=27$ times. Therefore, the sum of the digits in the units place is $27 \cdot(3+6+9)=486$. Similarly, in the tens, hundreds, and thousands places, we get the same sum of 486. In the end, the sum of all the numbers is

$$
486 \cdot 1000+486 \cdot 100+486 \cdot 10+486=486 \cdot 1111=539946
$$