<s>[INST] If $x\in\mathbb R$, solve $4x^2-40\lfloor x\rfloor+51=0$. 
If $x\in\mathbb R$, solve $$4x^2-40\lfloor x\rfloor+51=0$$
  where $\lfloor x\rfloor$ denotes the integer part of the number. $\lfloor x\rfloor\le x$ and $\lfloor x\rfloor=x-\{x\}$, where $\{x\}$ marks the fraction part of the number. $0\le\{x\}<1$.

Not sure if all these denotions are conventions that always mean what I've mentioned.
Also, I'm not too sure if $x\in\mathbb R$, since the problem doesn't mention it, but it seems quite obvious that it most likely is that way.
 [/INST] Note that when $i$ is odd then $3|(2^{i}-2)$ and hence $\lfloor \frac {2^{i}}{3} \rfloor = \lfloor \frac {2^{i}-2+2}{3} \rfloor = \lfloor \frac {2^{i}-2}{3} + \frac {2}{3} \rfloor = \frac {2^{i}-2}{3}$. 
Also when $j$ is even, then $3|(2^{j}-1)$ and hence $\lfloor \frac {2^{j}}{3} \rfloor = \lfloor \frac {2^{j}-1+1}{3} \rfloor = \lfloor \frac {2^{j}-1}{3} + \frac {1}{3} \rfloor = \frac {2^{j}-1}{3}$.
Therefore,
$$ \sum_{r=0}^{2010} \lfloor \frac {2^{r}}{3} \rfloor = \sum_i \lfloor \frac {2^{i}}{3} \rfloor + \sum_j \lfloor \frac {2^{j}}{3} \rfloor = \sum_i \frac {2^{i}-2}{3} + \sum_j  \frac {2^{j}-1}{3}$$
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