Jamaica
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^@
- 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^@. - 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. - 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^@. - 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^@.
- Hence, the maximum value of ^@n^@ is ^@18^@.
