Question

If $F(n)$ denotes the set of all divisors of $n$ except $1$. What is the least value of $y$ satisfying $[F(20)\cap F(16)]\subseteq F(y)$ ?
A. \[1\]
B. \[2\]
C. \[4\]
D. \[8\]

Answer 1

Hint: To solve this question we will first calculate $F(20)$ by collecting all the elements which are divisor of $20$ except $1$. Then we will calculate $F(16)$ in which all the elements will be divisor of $16$ except $1$. We will then calculate the intersection of $F(20)$ and $F(16)$ by selecting the elements which are similar in both sets. As $F(20)\cap F(16)$ is subset of $F(y)$, we will get the value of $F(y)$. Then we will count the number of elements which will be present in $F(y)$ thereby getting the least number of $y$ satisfying $[F(20)\cap F(16)]\subseteq F(y)$.



Complete step by step solution:We are given that $F(n)$ is the set of all divisors of $n$ except$1$ and we have to find the least value of $y$ which will satisfy $[F(20)\cap F(16)]\subseteq F(y)$.
We will first calculate the value of $F(20)$ in which all the elements will be divisor of $20$ except $1$.
$F(20)=\{2,4,5,10,20\}$
Now we will calculate the value of $F(16)$ in which all the elements will be divisor of $16$ except $1$.
$F(16)=\{2,4,8,16\}$
We know that the union of two sets contains all the elements of both the sets. So we will now calculate the union of $F(20)$ and $F(16)$ that is $F(20)\cap F(16)$.
$\begin{align}
  & F(20)\cap F(16)=\{2,4,5,10,20\}\cap \{2,4,8,16\} \\
 & =\{2,4\}
\end{align}$
As $F(20)\cap F(16)$ is subset of $F(y)$ , $F(y)$ will also contain the elements which are present in $F(20)\cap F(16)$ so we can say that $F(y)=\{2,4\}$ containing only two elements.
Hence the least value of$y$which will satisfy $[F(20)\cap F(16)]\subseteq F(y)$ is $2$.



Option ‘B’ is correct



Note: Subset can be defined as a part of a set. It means that if set A is subset of set B then elements of set A will be also present in set B. It is denoted by the symbol “$\subseteq $” which is read as “is a subset of “ and expressed as $P\subseteq Q$ which means that set P is a subset of set Q. A subset of any set can have all the elements of the set including null set.