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Question: If set of factors of a whole number n including n itself but not 1 is denoted by \(F\left( n \right)...

If set of factors of a whole number n including n itself but not 1 is denoted by F(n),F(16)F(40)=F(X)F\left( n \right),F\left( 16 \right)\cap F\left( 40 \right)=F\left( X \right) , then X is
(a) 4
(b) 8
(c) 6
(d) 10

Explanation

Solution

First, we have to make the list of factors of number X i.e. 16 and 40. So, we will find factors excluding digit 1. Thus, we will get 16=\left\\{ 2,4,8,16 \right\\} and 40=\left\\{ 2,4,5,8,10,20,40 \right\\} . Then, we will do intersection of these two lists i.e. we will take common numbers from the list and the number which will be highest in the list that will be F(X)F\left( X \right) . Thus, we will get an answer.

Complete step-by-step answer :
Here, we are told that F(n)F\left( n \right) is the set of whole numbers excluding 1 from its factor and including n as factor. So, first we will find factors of F(16),F(40)F\left( 16 \right),F\left( 40 \right) i.e. 16, 40.
So, we get as
16=(2×8),(4×4),(8×2),(16×1)16=\left( 2\times 8 \right),\left( 4\times 4 \right),\left( 8\times 2 \right),\left( 16\times 1 \right)
So, we can say that 16 is divisible by numbers like 2, 4, 8, 16. We can write is as
16=\left\\{ 2,4,8,16 \right\\} …………………………………(1)
Similarly, we will find 40. So, we get as
40=(2×20),(4×10),(5×8),(8×5),(10×4),  (20×2),(40×1) \begin{aligned} & 40=\left( 2\times 20 \right),\left( 4\times 10 \right),\left( 5\times 8 \right),\left( 8\times 5 \right),\left( 10\times 4 \right), \\\ & \text{ }\left( 20\times 2 \right),\left( 40\times 1 \right) \\\ \end{aligned}
So, we can here say that 40 is divisible by 2, 4, 5, 8, 10, 20, 40. Thus, factors of 40 as written as
40=\left\\{ 2,4,5,8,10,20,40 \right\\} ………………………………(2)
Now, we are given that
F(16)F(40)=F(X)F\left( 16 \right)\cap F\left( 40 \right)=F\left( X \right)
So, from equation (1) and (2), we will find which numbers are common in both that will be our answer. We can write it as
\left\\{ 2,4,8,16 \right\\}\cap \left\\{ 2,4,5,8,10,20,40 \right\\}
Thus, intersection points are
\left\\{ 2,4,8 \right\\}
So, here the last number is 8 therefore, we can get F(X)=F(8)F\left( X \right)=F\left( 8 \right) .
Thus, option (b) is the correct answer.

Note :Remember that here it is told that the factors of number which we want to find should be in the list i.e. 40=\left\\{ 2,4,5,8,10,20,40 \right\\} we can see that 40 is the number which we have to find factors and also 40 is in list so, here after finding intersection we got \left\\{ 2,4,8 \right\\} . So, we will consider that this list is for factor of number 8 and therefore the answer selected is 8. Do not select 4 here otherwise it is wrong. So, be careful while solving this type of problem.