Sum Quadratic Floor

The total number of terms is $66$ ,let us assume that $A$ of them have value $y$ and $B$ of them have value $(y+1)$ ,then, $Ay+By+B=625\\ \Longrightarrow 66y+B=625\\ \text{we also have that}\ B<66\\ \text{hence}\ y=9\ \text{and}\ B=31 \text{and}\ A=35$ .This means that the first $35$ values are $9$ and the $(36^{th}-66^{th})$ terms have a value of $10$ $\Longrightarrow [x+0.5]=9,\text{but}\ [x+0.51]=10\\ \Longrightarrow x=9.49.....\\ \Longrightarrow [100x]=949$ .And done!

Let the integer and fractional parts of $x$ be $\lfloor x \rfloor = n$ and $\{ x \} = \delta$ respectively. Then, we have:

$\begin{aligned} \sum_{k=16}^{81} \left \lfloor x + \frac{k}{100} \right \rfloor & = 625 \\ \Rightarrow \sum_{k=16}^{81} \left \lfloor n + \delta + \frac{k}{100} \right \rfloor & = 625 \\ \sum_{i=1}^{66} n + \sum_{k=16}^{81} \left \lfloor \frac{k}{100} + \delta \right \rfloor & = 625 \\ 66n + \sum_{k=16}^{81} \left \lfloor \frac{k}{100} + \delta \right \rfloor & = 625 \quad \color{#3D99F6} { \Rightarrow 66n \le 625 \quad \Rightarrow n = \left \lfloor \frac{625}{66} \right \rfloor = 9} \\ \color{#3D99F6}{594} + \sum_{k=16}^{81} \left \lfloor \frac{k}{100} + \delta \right \rfloor & = 625 \\ \Rightarrow \sum_{k=16}^{81} \left \lfloor \frac{k}{100} + \delta \right \rfloor & = 31 \\ \Rightarrow \sum_{k=\color{#3D99F6}{51}}^{81} 1 & = 31 \\ \color{#3D99F6}{\Rightarrow \left \lfloor \frac{51}{100} + \delta \right \rfloor} & \color{#3D99F6}{= 1\quad \Rightarrow 0.49 \le \delta < 0.5} \\ \color{#3D99F6}{\Rightarrow \left \lfloor 100x \right \rfloor} & \color{#3D99F6}{= \left \lfloor 100(n + \delta) \right \rfloor = \boxed{949}} \end{aligned}$

Since there are 81-16+1=66 terms, if x=9, we would add up to 594. We are 625-594=31 short. So last 31 terms ONLY must add up to 10. Last 31st term is 81-31 +1=51st, that is k=51. So we add $\dfrac {51}{100}=.51$ to x to make 10. So~ x=10-.51=9.49. ~~\Check for k=50 ;- $~~\lfloor 9.49+\dfrac {50}{100} \rfloor =\lfloor 9.99 \rfloor$ =9. So it is OK.