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

Hint: Assume ‘N’ as any integer lying between 1 and 102. Write the expression for the division of N by \[\sqrt{2}\] and rationalize the denominator. Use the property that “the product of a rational and an irrational number is always irrational”. Finally, use the property that “an irrational number is always non – repeating and non - terminating” to find the answer.

Complete step-by-step answer:
Here, we have been given integers lying between 1 and 102 and we have to find the total number of such integers which are divisible by \[\sqrt{2},\sqrt{3}\] and \[\sqrt{6}\].
Clearly, we can see that \[\sqrt{2},\sqrt{3}\] and \[\sqrt{6}\] are irrational numbers. An irrational number is a subset of real numbers that cannot be expressed as the ratio of two integers. An irrational number is both non – repeating and non – terminating. For example: - \[\sqrt{2}=1.414....,\sqrt{3}=1.732....\] and so on.
Now, let us consider a number ‘N’ that lies between 1 and 102 and it is an integer. Suppose N = 38. So, we have to determine whether 38 is divisible by \[\sqrt{2},\sqrt{3}\] and \[\sqrt{6}\] or not. Let us consider \[\sqrt{2}\] first, so we have,
\[\Rightarrow \] Expression = \[\dfrac{N}{\sqrt{2}}=\dfrac{38}{\sqrt{2}}\]
Rationalizing the denominator, we get,
\[\Rightarrow \] Expression = \[\dfrac{38}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{38\sqrt{2}}{2}=19\sqrt{2}\]
Now, we know that 19 is a rational number and \[\sqrt{2}\] is an irrational number and the product of a rational and an irrational number. Therefore, \[19\sqrt{2}\] will be an irrational number which will be non – repeating and non – terminating. Hence, \[\sqrt{2}\] will never divide 38.
Similarly, \[\sqrt{2}\] will not divide any integer and similar is the case with \[\sqrt{3}\] and \[\sqrt{6}\]. So, there are no integers that are divisible by \[\sqrt{2},\sqrt{3}\] and \[\sqrt{6}\].

So, the correct answer is “Option (b)”.

Note: Note that here we do not have to check the divisibility of each and every integer from 1 to 102. One example is enough to conclude the result. You must remember the properties of irrational numbers. You can see that we have rationalized the expression. This was done to make the observation easy although it was not necessary.