A) 99
B) 98
C) 66
D) 65
E) 33
Correct Answer: D
Solution :
\[\left[ \frac{2}{3}+\frac{r}{99} \right]=\left\{ \begin{matrix} 0 & x<33 \\ 1 & r\ge 33 \\ \end{matrix} \right.\] \[\therefore \] \[\sum\limits_{r=0}^{98}{\left[ \frac{2}{3}+\frac{r}{99} \right]}=\sum\limits_{r=0}^{32}{\left[ \frac{2}{3}+\frac{r}{99} \right]}\] \[=\sum\limits_{r=33}^{98}{\left[ \frac{2}{3}+\frac{r}{99} \right]}=0+65=65\]
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