7.2. Find the sum of all three-digit natural numbers that do not contain the digit 0 or the digit 5.
Answer: 284160. Solution. We will add the numbers in a column. Each last digit appears in the units place as many times as there are three-digit numbers ending with this digit. Therefore, it will appear $8 \cdot 8=64$ times (since a total of 8 digits are used for the hundreds and tens places). Thus, the sum of the digits in the last place is $64 \cdot(1+2+3+4+6+7+8+9)=2560$. Similarly, in the tens and hundreds places, we get the same sum. In the end, we get $2560 \cdot 100+2560 \cdot 10+2560=2560 \cdot 111=284160$.