4. Given a natural number $N$. Vera performs the following operations with it: first, she adds 3 until the resulting number is divisible by 5 (if $N$ is initially divisible by 5, nothing needs to be added). The resulting number is then divided by 5. She then performs the same operations with the new number, and so on. From which numbers can 1 not be obtained by these operations?
Solution. These are numbers of the form $3 k, k \in \mathbb{N}$. Indeed, from numbers that are multiples of 3, the number 1 cannot be obtained, because if a number $N$ is a multiple of 3, then $N+3$ is also a multiple of 3, and if $N=5 k$ is a multiple of 3, then $k$ is a multiple of 3, since 3 and 5 are coprime. Therefore, all numbers resulting from these operations will be multiples of 3, but 1 is not divisible by 3.

Let $N$ not be a multiple of 3. Notice that the numbers $N, N+3, N+6, N+9$, and $N+12$ are also not multiples of 3 and have different remainders when divided by 5, so one of them is a multiple of 5.

Therefore, after the first operation of dividing by 5, Vera will get a number that is not a multiple of 3 and does not exceed $\frac{N+12}{5}=0.2 N+2.4$, which is strictly less than $N$ for $N>3$. In other words, after each division by 5, these numbers decrease until the number 1 or 2 is obtained. But from the number 2, the number 1 can also be obtained in two steps.