Question

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

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

Answer 1

The correct option is C $66$

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

$[\frac{2}{3}+\frac{1}{99}]=0$

$[\frac{2}{3}+\frac{2}{99}]=0$

$...$

$...$

$[\frac{2}{3}+\frac{33}{99}]=[\frac{2}{3}+\frac{1}{3}]=\left[1\right]=1$

$[\frac{2}{3}+\frac{34}{99}]=[1+\frac{1}{99}]=1$

$[\frac{2}{3}+\frac{98}{99}]=[1+\frac{65}{99}]=1$

So, the sum is

$1+1+1+.........66$ times

$=66$