<s>[INST] Find number of points of discontinuity of $f(x)=\frac x5-\left\lfloor \frac x5\right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$ for $x\in[0,100]$ Find number of points of discontinuity of $$f(x)=\frac{x}{5}-\left\lfloor \frac{x}{5} \right\rfloor+\left\lfloor \frac{x}{2} \right\rfloor$$ for $x\in[0,100]$ .
My Attempt
The floor function is discontinuous for all integers .Now $\frac{x}{5}$ becomes integers at $21$ values of $x$ for $x\in[0,100]$ and $\frac{x}{2}$ becomes integer at $51$ values of $x$ for $x\in[0,100]$. So $f(x)$ should be discontinuous at $72$ values of $x$. But here the values of $x$ which are multiples of $10$ get counted twice so actually the function becomes discontinuous at $61$ values of $x$.
Given Answer
But in the answer given it was said that $f(x)$ is continuous at such values of $x$ which are multiples of $10$.I checked and found it to be true. So the function can be said to be discontinuous at $61-11=50$ points.
My Issue
How is one supposed to know for sure that $f(x)$  will be continuous at such $x$ which are multiples of $10$
 [/INST] $\left\lfloor \frac{x}{5}\right\rfloor$ is continuous in the multiples of $5$.
$\left\lfloor \frac{x}{2}\right\rfloor$ is continuous in the multiples of $2$. So $f$ is continuous in the integers that are simultaneously multiples of $5$ and $2$, since $\gcd(2,5)=1$, those are the multiples of $\operatorname{lcm}(2,5)=10$.
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