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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!

 Jan 22, 2022
 #1
avatar+31 
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What is your solution to this problem? I'm curious.

 Jan 23, 2022
 #2
avatar+117100 
+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

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