9.1 How many five-digit natural numbers are there that are divisible by 9, and for which the last digit is 2 more than the second last digit?
Answer: 800. Hint The second last digit (the tens digit) can be any from 0 to 7 (so that after adding two, the last digit makes sense). The third and second digits can be any (from 0 to 9). After choosing the specified three digits (the second, third, and fourth), the last digit is uniquely determined by the second last, and the first digit is uniquely determined by the other four under the condition of divisibility by 9 (note that the first digit cannot be 0, and all other digits $1,2, \ldots, 9$ give all possible remainders when divided by 9). In total, by the rule of product, we have $8 \cdot 10 \cdot 10=800$ numbers.