IPMAT Question Paper 2019 | IPM Indore Quants

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Question 14 : Let the set = {2,3,4,..., 25}. For each k ∈ P, define Q(k)= {x ∈ P such that x > k and k divides x}. Then the number of elements in the set \ P - U_{k=2}^{25} \\) Q(k) is

Explanatory Answer

Let’s take a close look at the definition of Q(k).

Q (k) = {x ∈ P such that x > k and k divides x}.
That means Q(k) contains all the multiples of k in P which are greater than k.

\ U_{k=2}^{25} \\) Q(k) means Q(2) U Q(3) U Q(4) U ………… U Q(25).

Therefore, Q(2) U Q(3) U Q(4) U ………… U Q(25) will include every composite number in P.

Because every composite number is a multiple of at least one prime number lesser than itself.

For Example, 21: Q(3) contains 21, Q(7) contains 21.
So, 21 is surely present once in Q(2) U Q(3) U Q(4) U ………… U Q(25) ( once and only once, because repetitions are neglected in Union)

But Q(2) U Q(3) U Q(4) U ………… U Q(25) will not include prime numbers in P at all.

Because, a prime number can never be expressed as a multiple of another number except 1, which is not present in P.

So we can conclude that Q(2) U Q(3) U Q(4) U ………… U Q(25) contains only composite numbers.
That is, \ U_{k=2}^{25} \\) Q(k) contains only composite numbers.

Therefore, \ P - U_{k=2}^{25} \\) Q(k) = P – {composite numbers in P} = {prime numbers in P} ={2,3,5,7,11,13,17,19,23}

So, the number of elements in the set \ P - U_{k=2}^{25} \\) Q(k) = the number of elements in the set {2,3,5,7,11,13,17,19,23} = 9.

The number of elements in the set \ P - U_{k=2}^{25} \\) Q(k) is 9.

The question is "Let the set = {2,3,4,..., 25}. For each k ∈ P, define Q(k)= {x ∈ P such that x > k and k divides x}. Then the number of elements in the set \ P - U_{k=2}^{25} \\) Q(k) is"

Hence, the answer is 9

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