7.2. For a 92-digit natural number n, the first 90 digits are known: from the 1st to the 10th - ones, from the 11th to the 20th - twos, and so on, from the 81st to the 90th - nines. Find the last two digits of the number n, given that n is divisible by 72.
Answer: 36. Solution: Let the last digits be $x$ and $y$. The number $n$ must be divisible by 9 and 8. The number consisting of the first 90 digits is divisible by 9, since the sum of its digits is divisible by 9. Therefore, the number $\overline{x y}$ is also divisible by 9. In addition, by the divisibility rule for 4, the number $\overline{x y}$ is divisible by 4. Therefore, $\overline{x y}$ is either 00, 36, or 72. Since $n$ is divisible by 8, the number $\overline{9 x y}$ (consisting of the last three digits) is divisible by 8. Among the numbers 900, 936, and 972, only 936 has this property.