Question

Number of integers lying between 1 to 102 which are divisible by all $\sqrt{2},\sqrt{3},\sqrt{6}$ is

• A. $16$
• B. $17$
• C. $15$
• D. $0$

Answer 1

The correct option is D $0$

For a number to be divisible by $\sqrt{2}$, it must be an irrational number.

An integer is not an irrational number.

So there are no numbers between $1$ to $102$ which all $\sqrt{2},\sqrt{3},\sqrt{6}.$

So option D is the correct answer.