Find the smallest number that is not divisible by 27 itself, but its sum of digits is divisible by 27.

Obviously, there cannot be such a number among single-digit and two-digit numbers (the maximum single-digit number is 9, and the maximum two-digit number is 99 with the sum of 18 digits).
Among three-digit numbers, only one number can have the sum of 27 digits; this number is 999. Let’s factor it out into prime factors: 999 = 3 * 3 * 3 * 37. This number is divisible by 27.
Go ahead. Among the four-digit numbers, the smallest number with the sum of digits equal to 27 is the number 1899. It is easy to compose: as you know, the smallest digit is 0, however, it cannot be the initial digit of a natural number. Then, as the first digit, we take 1. In this case, the remaining 3 digits must add up to the sum of digits equal to 27 – 1 = 26. It is clear that the minimum such number is 899.
So, let’s check the number 1899 for divisibility by 27. To do this, factor the number 1899. Since the sum of the digits is divisible by 3, then 1899 = 3 * 633. The sum of the digits of the number 633 is 12, which is divisible by 3. So, 1899 = 3 * 3 * 211. Checks for signs of divisibility by initial primes 3, 5, 7, 11, 13 and 17 force us to refer to the table of the first 100 primes. It turns out that 211 is a prime number.
This means that 1899 is not a multiple of 27.
Answer: The smallest number that itself is not divisible by 27, but its sum of digits is divisible by 27, is 1899.