Solve the equation $[x / 2]+[x / 4]=[x]$. ( $([x]$, the integer part of $x$ is the largest integer not greater than $x$.)
I. solution. Let's confine both sides of the equation within bounds. Based on the definition of the integer part:

$$
\frac{3}{4} x-2<\left[\frac{x}{2}\right]+\left[\frac{x}{4}\right] \leq \frac{3}{4} x, \quad x-1<[x] \leq x
$$

If $\frac{3}{4} x \leq x-1$, that is, $4 \leq x$, or $\frac{3}{4} x-2 \geq x$, that is, $-8 \geq x$, then the equation has no solution. Thus, the solution must lie in the remaining $(-8 ; 4)$ interval. Divide this interval into smaller parts where both sides of the equation are constant; these values are contained in the following table:

| interval | $\left[\frac{x}{2}\right]+\left[\frac{x}{4}\right]$ | $[x]$ |  |
| :--- | :---: | :---: | :--- |
| $(-8 ;-7)$ | -6 | -8 |  |
| $[-7 ;-6)$ | -6 | -7 |  |
| $[-6 ;-5)$ | -5 | -6 |  |
| $[-5 ;-4)$ | -5 | -5 | (solution) |
| $[-4 ;-3)$ | -3 | -4 |  |
| $[-3 ;-2)$ | -3 | -3 | (solution) |
| $[-2 ;-1)$ | -2 | -2 | (solution) |
| $[-1 ; 0)$ | -2 | -1 |  |
| $[0 ; 1)$ | 0 | 0 | (solution) |
| $[1 ; 2)$ | 0 | 1 |  |
| $[2 ; 3)$ | 1 | 2 |  |
| $[3 ; 4)$ | 1 | 3 |  |

Thus, the solutions of the equation are the intervals $[-5 ;-4),[-3 ;-1),[0 ; 1)$. () Tamás Bitai (Bonyhád, Petőfi Sándor Evangelical Gymnasium, 10th grade)

II. solution. If $[x]=n$, then $\left[\frac{x}{2}\right]=\left[\frac{n}{2}\right]$ and $\left[\frac{x}{4}\right]=\left[\frac{n}{4}\right] ; x$ is thus a solution to the problem if and only if

$$
n=\left[\frac{n}{2}\right]+\left[\frac{n}{4}\right]
$$

We need to examine four cases depending on whether the number $n$ can be written in the form $4 k, 4 k+1, 4 k+2$, or $4 k+3$ for some integer $k$.

In the first case, our equation is in the form $4 k=2 k+k$, from which $k=0, n=0$.

In the second case, it is in the form $4 k+1=2 k+k$, from which $k=-1, n=-3$.

In the third case, it is in the form $4 k+2=(2 k+1)+k$, from which $k=-1, n=-2$.

In the fourth case, it is in the form $4 k+3=(2 k+1)+k$, from which $k=-2, n=-5$.

Thus, the solutions to the problem are those real numbers $x$ whose integer part is $0, -3, -2$, or $-5$, that is, the solution set of the equation is: $[-5 ;-4) \cup[-3 ;-1) \cup[0 ; 1)$.