### Let $S = \{1, 2, 3, ....., 46, 47\}$. What is the maximum value of $n$ such that it is possible to select $n$ numbers from $S$ and arrange them in a circle in such a way that the product of any two adjacent numbers in the circle is less than $100?$

$18$

Step by Step Explanation:
1. Given $S = \{1, 2, 3, ....., 46, 47\}$.
Now, we know that the product of any two $2$-digit numbers is either equal to or more than $100$.
2. If $n$ numbers are chosen from $S$ and arranged as per the question, no two $2$-digit numbers are adjacent.
Therefore, the two numbers adjacent to a $2$-digit number must be single-digit numbers.
3. A maximum of nine $1$-digit numbers can be chosen from $S$ and a $2$-digit number can fit in between any two $1$-digit numbers. There will be $9$ such places between any two $1$-digit numbers.
Without loss of generality, let us place the numbers $1, 2, 3,...,9$ in the ascending order. Now place the numbers $10, 11, 12, ...,18$ in between these numbers such that $18$ is placed between $1$ and $2, 17$ is placed between $2$ and $3, 16$ is placed between $3$ and $4$ and so on to ensure that the product of any two adjacent numbers is less than $100$.
4. Therefore, one can choose a maximum of $18$ numbers $($nine $1$-digit numbers and nine $2$-digit numbers$)$ from $S$ and arrange them in a circle in such a way that the product of any two adjacent numbers in the circle is less than $100$.
5. Hence, the maximum value of $n$ is $18$.