Question

Assertion :For $xϵR$ ,let [x] denote the greatest integer $\le x$,then the number A defined by $A=\left[\frac{1}{3}\right]+[\frac{1}{3}+\frac{1}{100}]+.......[\frac{1}{3}+\frac{99}{100}]$ is divisible by exactly two primes Reason: $[x+n]=n+\left[x\right],n\in I,$ & $[y+z]=\left[y\right]+\left[z\right]$ if $y,z\in I$

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A),
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A)is true but (R} is false,
• D. (A)is false but (R} is true.

Answer 1

The correct option is A Both (A) & (R) are individually true & (R) is correct explanation of (A),

$\frac{2}{3}=\mathrm{0.6666...}=\frac{\mathrm{66.666...}}{100}$

$A=\left[\frac{1}{3}\right]+[\frac{1}{3}+\frac{1}{100}]+......+[\frac{1}{3}+\frac{66}{100}]+\left[\frac{1}{3}\frac{67}{100}\right]+....+\left[\frac{1}{3}\frac{99}{100}\right]$

$=(0+0+......66times)+(1+1+1+.....33times)=33A=3\times 11$

$\therefore$ A has exactly two primes, so Assertion (A) -is true and Reason is obvious with proper (Correct) explanation