How many positive integers between 1 and 1988 can be expressed in the form

$$
[x]+[2 x]+[4 x]+[8 x]
$$

for some real number $x$?
Let $f(x)=[x]+[2 x]+[4 x]+[8 x]$. Since $x=[x]+\{x\}$, for any integer $n$,

$$
[n x]=[n[x]+n\{x\}]=n[x]+[n\{x\}]
$$

thus

$$
f(x)=15[x]+f(\{x\})
$$

Let's examine what values $f$ takes in the interval $[0,1)$! It is easily observable that $f(x)$ in this case depends only on which of the eight equal parts of the interval $x$ falls into. The function values can be summarized in the following table:

| $x \epsilon$ | $\left[0, \frac{1}{8}\right)$ | $\left[\frac{1}{8}, \frac{2}{8}\right)$ | $\left[\frac{2}{8}, \frac{3}{8}\right)$ | $\left[\frac{3}{8}, \frac{4}{8}\right)$ | $\left[\frac{4}{8}, \frac{5}{8}\right)$ | $\left[\frac{5}{8}, \frac{6}{8}\right)$ | $\left[\frac{6}{8}, \frac{7}{8}\right)$ | $\left[\frac{7}{8}, 1\right)$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $f(x)$ | 0 | 1 | 3 | 4 | 7 | 8 | 10 | 11 |

Thus, the set of values of the function $f$ consists precisely of numbers of the form $15 k+r$, where $k$ is an integer, and $r$ is one of the numbers $0,1,3,4,7,8,10,11$.

Since $|r|<15$, such numbers can be written in this form in only one way; thus, we need to count how many such $(k, r)$ pairs satisfy

$$
1 \leqq 15 k+r \leqq 1988
$$

$1988=132 \cdot 15+8$, so $0 \leqq k \leqq 132$. If $1 \leqq k \leqq 131$, then $r$ can take all 8 values; for $k=0$, only $r=0$ is valid, and for $k=132$, only $r=10$ and $r=11$ result in a number outside the interval $[1,1988]$. Therefore, the function $f$ has $133 \cdot 8-3=1061$ values between 1 and 1988.