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?


Answer:

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.

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