For which real numbers $x$ is it true that

$$
\left\{\frac{1}{3}\left[\frac{1}{3}\left(\frac{1}{3} x-3\right)-3\right]-3\right\}=0
$$

where $\{z\}$ is the fractional part of $z$ - see the 398th page of our November 1985 issue for the Gy. 2294 exercise -, $[z]$ is the integer part of $z$, i.e., $[z]=z-\{z\}$.
When solving the equation, it is necessary to use the fact that the fractional part of a number is exactly 0 if and only if the number is an integer, and that the integer part of a number $t$ is exactly equal to an integer $m$ if and only if $m \leq t < m+1$.

The solutions of the equation are then those real numbers $x$ for which

$27k + 9 \leq x < 27k + 18$, where $k$ is any integer.

Many did not notice that an integer part symbol also appears between the nested parentheses. Many concluded their exams by stating that "the correctness of the roots can be verified by substitution." Well, in the case of infinitely many roots, there is no way to do this; in such cases, we need to reconsider whether our steps are reversible and if so, we should refer to this.