<s>[INST] Find the total number of different integers that the function takes 
Find the total number of different integers that the function $$f(x) = [x]+[2x]+\left[\dfrac{5x}{3}\right]+[3x]+[4x]$$ takes for $0 \leq x \leq 100$. (Note: $[x]$ denotes the greatest integer not exceeding $x$.)

Is my solution below correct?
Attempt:
We see that the number of integer values of $f(x)$ where $x \in [0,3)$ is the same as the number of integer values of $f(x)$ where $x \in [3,6)$ and so on. The number of different integer values of $f(x)$ where $x \in [0,3)$ is equal to the number of places $f(x)$ changes value from where it was before plus one (its initial starting position at $x = 0$). Thus, the places where $f(x)$ changes its value can occur only when, expressed in simplest form, the numerator is a multiple of $3$ or when the denominator is a multiple of $3,4,$ or $5$. We then obtain the following $22$ fractions: $$0,1,2,\dfrac{1}{2},\dfrac{3}{2},\dfrac{5}{2},\dfrac{1}{3},\dfrac{2}{3},\dfrac{4}{3},\dfrac{5}{3},\dfrac{7}{3},\dfrac{8}{3},\dfrac{1}{4},\dfrac{3}{4},\dfrac{5}{4},\dfrac{7}{4},\dfrac{11}{4},\dfrac{3}{5},\dfrac{6}{5},\dfrac{9}{5},\dfrac{12}{5}.$$ Similarly, in the interval $[99,100)$ we see that are $8$ integer values of $f(x)$ generated here. Thus, there are $33 \times 22+8 = 734$ different integer values of $f(x)$ taken on this range.
 [/INST] Your approach is correct, though your logic is written in a way that seems somewhat ambiguous. The most notable thing is that the condition for $f$ to experience a jump at $x$ is that $x$ is a (integer) multiple of  one of $1,\,\frac{1}2,\,\frac{1}3,\,\frac{1}4,\,\frac{3}5$. If you need more formality, this a convenient formulation, since you can prove that each summand of the form $[ax]$ is constant except where $x=\frac{c}a$ for some integer $c$. You can conclude that $f(y)$ is different from $f(x)$ exactly when there is some number of the form $\frac{c}a$ between them for $a$ being one of $1,\,2,\,3,\,4,\,\frac{5}3$ - i.e. the given coefficients.
Note that this has little to do with the divisibility of the numerator or denominator. Your current phrasing suggests that $\frac{1}8$ should be included in the list, as the denominator is a multiple of $4$, and that $\frac{3}7$ should be included in the list, as the numerator is a multiple of $3$. While your list is correct, the criteria you used for inclusion in it is different than what you stated.
The other small detail in your proof that seems suspicious is where you talk about "the number of places $f(x)$ changes value from where it was before plus one" - it seems to imply that all the jumps in $f(x)$ have a height of $1$, however, one can note that the jumps in the constituent functions overlap. For instance, $f(x)$ takes the value of $30$ when $x$ is in $[3-1/5,3)$, but jumps to $35$ at $3$, creating a jump with height $5$ due to every constituent function's jump lining up.
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