Find the total number of different integer values the function

$$
f(x)=[x]+[2 x]+\left[\frac{5 x}{3}\right]+[3 x]+[4 x]
$$

takes for real numbers \( x \) with \( 0 \leq x \leq 100 \).
Note: \([t]\) is the largest integer that does not exceed \( t \).
Answer: 734.
Note that, since $[x+n]=[x]+n$ for any integer $n$,

$$
f(x+3)=[x+3]+[2(x+3)]+\left[\frac{5(x+3)}{3}\right]+[3(x+3)]+[4(x+3)]=f(x)+35,
$$

one only needs to investigate the interval $[0,3)$.
The numbers in this interval at which at least one of the real numbers $x, 2 x, \frac{5 x}{3}, 3 x, 4 x$ is an integer are

- $0,1,2$ for $x$;
- $\frac{n}{2}, 0 \leq n \leq 5$ for $2 x$;
- $\frac{3 n}{5}, 0 \leq n \leq 4$ for $\frac{5 x}{3}$;
- $\frac{n}{3}, 0 \leq n \leq 8$ for $3 x$;
- $\frac{n}{4}, 0 \leq n \leq 11$ for $4 x$.

Of these numbers there are

- 3 integers $(0,1,2)$;
- 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );
- 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );
- 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );
- 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).

Therefore $f(x)$ increases 22 times per interval. Since $100=33 \cdot 3+1$, there are $33 \cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \frac{1}{2}, 99 \frac{1}{3}, 99 \frac{2}{3}, 99 \frac{1}{4}$, $99 \frac{3}{4}, 99 \frac{3}{5}$.
The total is then $33 \cdot 22+8=734$.
Comment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of

$$
0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30
$$

in the interval $[0, f(100)]=[0,1166]$. Since $1166 \equiv 11(\bmod 35)$, this comprises 33 cycles plus the 8 numbers in the previous list.