Adding Up A Lot Of Floors

$\begin{aligned} S & = \left \lfloor x+\frac{1}{100} \right \rfloor + \left \lfloor x+\frac{2} {100} \right \rfloor +\ldots + \left \lfloor x+\frac{223}{100} \right \rfloor=521 \\ &= 3\left( \left \lfloor x+\frac{1}{100} \right \rfloor + \left \lfloor x+\frac{2}{100} \right \rfloor + \ldots +\left \lfloor x+\frac{23}{100} \right \rfloor \right)+2 \left( \left \lfloor x+\frac{24}{100} \right \rfloor + \left \lfloor x+\frac{25}{100} \right \rfloor + \ldots +\left \lfloor x+\frac{100}{100} \right \rfloor \right) =521-100-23×2=375 \\ & \Rightarrow x = \lfloor x\rfloor+ \{x\} \\ S & = 223\lfloor x\rfloor + 3 \displaystyle \sum^{23}{n=1} \left \lfloor \{x\}+\frac{n}{100} \right \rfloor +2 \displaystyle \sum^{100}_{m=24} \left \lfloor \{x\}+ \frac{m}{100} \right \rfloor =375 \\ S &=375, \quad \lfloor x\rfloor = \left \lfloor \frac{375}{223} \right \rfloor =1 \\ & \Rightarrow 3 \displaystyle \sum^{23}_{n=1} \left \lfloor \{x\}+\frac{n}{100} \right \rfloor +2 \displaystyle \sum^{100}_{m=24} \left \lfloor \{x\}+ \frac{m}{100} \right \rfloor = 375-223 = 152 \end{aligned}$

Now we can obviously see that any term of $\left( \left \lfloor x+\frac{n}{100} \right \rfloor \right)^{23}_{n=1}$ would be smaller than any term of $\left(\left \lfloor x+\frac{n}{100} \right \rfloor \right)^{100}_{n=24}$ , so $152$ would coming from $2 \displaystyle \sum^{100}_{m=24} \left \lfloor \{x\}+ \frac{m}{100} \right \rfloor$ and note that the terms can't be greater than $1$ . If all the terms equal 1 , then $2 \displaystyle \sum^{100}_{m=24} \left \lfloor \{x\}+ \frac{m}{100} \right \rfloor =154$ , so to decrease $2$ and get $152$ as our result, all the terms of $\left( \left \lfloor \{x\} +\frac{m}{100} \right \rfloor \right)^{100}_{m=25}$ must be equal to $1$ .

$\begin{aligned} \therefore \{x\} + \frac{25}{100} & \geq 1 \\ \{x\} & \geq 0.75 \\ x & =\lfloor x \rfloor +\{x\}=1+0.75 =1.75 \\ \Rightarrow \lfloor 100x \rfloor & = \boxed{175} \end{aligned}$

Suppose 'k' of them be 1,100 of them be 2 and '123-k' be 3 So, k+200+3(123-k)=521 Solving we get k=24. So,[x+24/100]=1 but [x+25/100]=2 Thus, [100x]={2-(1/4)}100=175(ans.)