Calculate $\sum_{m=0}^{100}\left\lfloor \frac{24\cdot m}{101} \right\rfloor$ How can I calc this sum:  $$\sum_{m=0}^{100}\left\lfloor \frac{24\cdot m}{101} \right\rfloor$$ 
I stuck because the floor...
 A: Thank you, Gauss!
For integer $m$ from $1$ to $100$,
$$\left\lfloor \frac{24\cdot m}{101} \right\rfloor+\left\lfloor \frac{24\cdot (101-m)}{101} \right\rfloor=\frac{(24)(101)}{101}-1=23$$ because both $24m$ and $24(101-m)$ are relatively prime to $101$ (note: 101 is a prime). 
(more explanation: if $a$ and $b$ are not integers, but their sum is an integer, then $a+b$ is equal to the floor of $a$ plus the floor of $b$ minus one. This is because the decimal parts of $a$ and $b$ must add up to $1$ (it must be an integer. It cannot be 0 otherwise $a$ and $b$ will both be $0$. It cannot be $2$ since each of the fractional parts are less than $1$. 
Summing this for $m=1$, $2$, ..., $50$, we get the desired summation and the value is $(23)(50)=1150$. 
A: $1.$ Let $f(x) =  \frac{24\cdot m}{101} $. The minimum and maximum values that $24\cdot m$ can take are $0$ and $2400$.
$2.$ Now check the range $f(x)$. You can notice that the range lies between $0$ and $\frac{2400}{101}$.
$3.$ For $0 \leq m \leq 4$, $\frac{24\cdot m}{101} < 1$. Similarly, for $5 \leq m \leq 8$, $1 < \frac{24\cdot m}{101} < 2.$ Split the given expression into such intervals as it will be be easy to calculate the $\left\lfloor f(x) \right\rfloor$
