Question

If $\left[x\right]$ denotes the greatest integer $\le x$, then $\left[\frac{2}{3}\right]+\left[\left(\frac{2}{3}\right)+\left(\frac{1}{99}\right)\right]+\left[\left(\frac{2}{3}\right)+\left(\frac{2}{99}\right)\right]+\dots .+\left[\left(\frac{2}{3}\right)+\left(\frac{98}{99}\right)\right]=?$

• A. $99$
• B. $98$
• C. $66$
• D. $65$
• E. $33$

Answer 1

The correct option is C

$66$

Explanation for the correct option

$\left[\frac{2}{3}\right]=\left[0.67\right]=0$

$\left[\frac{2}{3}+\frac{r}{99}\right]$ will be an integer at $r=33$, where the value of $\left[\frac{2}{3}+\frac{r}{99}\right]=\left[1\right]=1$

We will get the next integer when, $\left[\frac{2}{3}+\frac{r}{99}\right]=2$, that is when $r=132$, which is more than $98$.

So, the value of the given greatest integer function, can be $0\mathrm{or}1$.

Now, $\left[\frac{2}{3}+\frac{r}{99}\right]$ will be $1$, when $98\ge r\ge 33$, and it will be $0$when $0\le r<33$

Therefore,

$\begin{array}{rcl}\left[\frac{2}{3}\right]+\left[\left(\frac{2}{3}\right)+\left(\frac{1}{99}\right)\right]+\left[\left(\frac{2}{3}\right)+\left(\frac{2}{99}\right)\right]+\dots .+\left[\left(\frac{2}{3}\right)+\left(\frac{98}{99}\right)\right]& =& 0+\sum _{r=33}^{98}\left[\frac{2}{3}+\frac{r}{99}\right]\\ & =& 0+\left[1\right]\times 66\left(\mathrm{as}98-32=66\right)\\ & =& 66\end{array}$

Hence, option C is correct.