Question

The value of the expression $\left[\frac{10}{27}\right]+\left[\frac{11}{27}\right]+\left[\frac{12}{27}\right]+...+\left[\frac{35}{27}\right]$ is

• A. 9.00
• B. 9
• C. 9.0

Answer 1

Answer: (A), (B), (C)

$\left[\frac{10}{27}\right]+\left[\frac{11}{27}\right]+\left[\frac{12}{27}\right]+...+\left[\frac{35}{27}\right]=[\frac{9}{27}+\frac{1}{27}]+[\frac{9}{27}+\frac{2}{27}]+\dots +[\frac{9}{27}+\frac{26}{27}]=[\frac{1}{3}+\frac{1}{27}]+[\frac{1}{3}+\frac{2}{27}]+\dots +[\frac{1}{3}+\frac{26}{27}]$

We know that,
$\left[x\right]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+\dots +[x+\frac{n-1}{n}]=\left[nx\right]\Rightarrow [x+\frac{1}{n}]+[x+\frac{2}{n}]+\dots +[x+\frac{n-1}{n}]=\left[nx\right]-\left[x\right]$
So,
$[\frac{1}{3}+\frac{1}{27}]+[\frac{1}{3}+\frac{2}{27}]+\dots +[\frac{1}{3}+\frac{26}{27}]=[27\times \frac{1}{3}]-\left[\frac{1}{3}\right]=\left[9\right]-0=9$