Hello! Here is a really fun problem I came up with.

You are given a perfect square \(n.\)

Define a *superpair* in \(n\) as a \(2(n-1) \)-tuple, \((x_1, x_2, x_3, ... x_{2(n-1)})\), *such that* \(x_1\cdot x_2 \cdots x_{n-1} = x_n \cdot x_{n+1} \cdots x_{2(n-1)},\) with \(1 \leq x_i \leq n\). In other words, all elements must be between \(1\) and \(n\) and the product of its first \(n-1\) elements equals the product of its last \(n-1\) elements.)

Find the set \(S\) of values of \(n\) that *guarantee* that it *is possible* to form list of \(n\) superpairs in \(n\) such that every number that *appears in* your superpair list appears exactly twice in your superpair list.

For example, we know that \(n=3\) is in \(S\) because we can satisfy the conditions with the following superpair list:

\((1,6,2,3)\)

\((2,4,1,8)\)

\((3,8,6,4)\)

Here, \(1\cdot 6 = 2 \cdot 3,\) \(2 \cdot 4 = 1 \cdot 8,\) and \(3 \cdot 8 = 6 \cdot 4.\) Additionally, every number present in the the superpair list, (\(1,2,3,4,6,8\)) appears exactly twice.

Notice that *not* all numbers from \(1\) to \(n\) *need* to be in your superpair list. Rather, you must only ensure that the ones that *are in your superpair list,* appear exactly twice. *Hint: *try using exactly \(n^2-n\) distinct numbers in your superpair list.

I hope you enjoy!

ibhar Jan 22, 2022

#1

#2**+1 **

"What is your solution to this problem? I'm curious."

"Hint: try using exactly n^2 -n distinct numbers in your superpair list.

How can you give us a hint if you do not know what the answer is ?

If you came up with it yourself and you do not know what the answer is then how can you claim it is 'beautiful'?

Melody
Jan 23, 2022