[ Classical combinatorics (miscellaneous).] $[\quad$ Case enumeration $\quad]$

Find all odd natural numbers greater than 500 but less than 1000, for each of which the sum of the last digits of all divisors (including 1 and the number itself) is 33.
For an odd number, all its divisors are odd. Since the sum of their last digits is odd, there must be an odd number of divisors. If a number has an odd number of divisors, then it is a square of a natural number (see problem $\underline{30365}$). Let's consider all the squares of odd numbers within the given range: $23^{2}, 25^{2}=625,27^{2}=729,29^{2}, 31^{2}$. Note that the number we are looking for must have at least five divisors; otherwise, the sum of the last digits of its divisors would not exceed 27. This means that we can exclude the squares of prime numbers 23, 29, and 31, which have exactly three divisors. Therefore, only two numbers remain to be checked. The divisors of the number $625: 1,5,25,125,625$; the sum of their last digits is 21. The divisors of the number $729: 1,3$, $9,27,81,243,729$; the sum of their last digits is 33.

## Answer

729.