imc

Since $x=\lfloor x\rfloor +\{x\}$, we have $$f(x)=\sum _{k=2}^{10}(\lfloor k\lfloor x\rfloor +k\{x\}\rfloor -k\lfloor x\rfloor )$$ The function can then be simplified into $$f(x)=\sum _{k=2}^{10}(k\lfloor x\rfloor +\lfloor k\{x\}\rfloor -k\lfloor x\rfloor )$$ which becomes $$f(x)=\sum _{k=2}^{10}\lfloor k\{x\}\rfloor$$ We can see that for each value of $k$, $\lfloor k\{x\}\rfloor$ can equal integers from $0$ to $k-1$. Clearly, the value of $\lfloor k\{x\}\rfloor$ changes only when $\{x\}$ is equal to any of the fractions $\frac{1}{k},\frac{2}{k}\dots \frac{k-1}{k}$. So we want to count how many distinct fractions less than $1$ have the form $\frac{m}{n}$ where $n\le 10$. Explanation for this is provided below. We can find this easily by computing $$\sum _{k=2}^{10}\varphi (k)$$ where $\varphi (k)$ is the Euler Totient Function. Basically $\varphi (k)$ counts the number of fractions with $k$ as its denominator (after simplification). This comes out to be $31$. Because the value of $f(x)$ is at least $0$ and can increase $31$ times, there are a total of $\boxed{32}$ different possible values of $f(x)$.