Black/square magic

[Kieran]

The smallest number in a magic square is 1 less than 3 times the largest in the square. What is the number ? Is it possible to solve without a computer program ? No idea of the size of the matrix. Is there a technique or is it a case of hit and miss ? If I knew the size of the grid, would it be much easier ? Thank you.

 

[lookagain]

The smallest number in a magic square is 1 less than 3 times the largest in the square.
What is the number ? Is it possible to solve without a computer program ? No idea of the size of the matrix. I
s there a technique or is it a case of hit and miss ? If I knew the size of the grid, would it be much easier ? Thank you.

Suppose that I am going to consider (for the time being) only those magic squares that have distinct, consecutive integers for their entries.

How many entries are in a magic square with your constraints?

The number of entries is equal to the largest number - the smallest number + 1.


Let n = the largest number

Then 3n - 1 = the smallest number


The number of entries =

n - (3n - 1) + 1 =

n - 3n + 1 + 1

-2n + 2


Now, this number of entries is necessarily even, so for the usual magic squares,
we can consider only those squares which are 4 by 4, 6 by 6, 8 by 8, and so on.


Set -2n + 2 = 16 for beginners.

n = -7


3n - 1 =

3(-7) - 1 =

-22

So, the 16 numbers in that 4 by 4 magic square range from -22 to -7, inclusive.


Then the smallest number for that case is -22.



-------------


What if it is a 6 by 6 one? There are 36 entries.

Set -2n + 2 = 36

-2n = 34

n = -17


3n - 1 =

3(-17) - 1 =

-52


From smallest to largest, the 36 entries range from -52 to -17, inclusive.

Then the smallest number in that magic square is -52.


------------------


There exist magic squares for every n > 2, giving n^2 number of entries.

http://en.wikipedia.org/wiki/Magic_Square




Therefore, there are an infinite number of answers as to what is the smallest number is,
given that the smallest number (entry) is one less than three times the largest number
(entry).

 

[soroban]

Hello, Kieran!

I assume that the Magic Square has these properties:
. . An n-by-n square contains n² consecutive integers.
. . All rows, columns, and the two main diagonals have the same sum.

I'm familiar with Magic Squares, so I found a solution.
I don't think I can explain my reasoning, though.

The smallest number in a magic square is 1 less than 3 times the largest in the square.
What is the number?

This can be true only if the numbers are negative.

I found a 4-by-4 Magic Square using the numbers from -22 to -7.

Is it possible to solve without a computer program?
No idea of the size of the matrix.
Is there a technique or is it a case of hit and miss?
If I knew the size of the grid, would it be much easier?

I happen to know a trick for a 4-by-4 Magic Square

. . \(\displaystyle \begin{array}{|c|c|c|c|} \hline \text{-}7&\text{-}21&\text{-}20&\text{-}10 \\ \hline \text{-}18&\text{-}12&\text{-}13&\text{-}15 \\ \hline \text{-}14&\text{-}16&\text{-}17&\text{-}11 \\ \hline \text{-}19&\text{-}9&\text{-}8&\text{-}22 \\ \hline \end{array}\)

Magic sum: \(\displaystyle \text{-}58\)

 

[HallsofIvy]

The smallest number in a magic square is 1 less than 3 times the largest in the square.

As Soroban said, this is only possible if the numbers in the square are negative. Did you intend that? Or did you mean "The largest number in a magic square is 1 less than 3 times the smallest in the square?

 

[lookagain]

Whatever you say is correct.

I will attribute this to belonging to the 5% actual help of posts that I give to users
(as you stated to Ted in one of the threads months ago.)



I think the intent of the problem was to have the largest number be one less than three
times the smallest number, as opposed to what was stated by the original poster.

N    S    F(irst)     L(ast)
1   16       8           23
2   36      18           53        
3   64      32           95
4  100      50          149      
5  144      72          215
6  196      98          243
          
...to infinity

F = 2(N + 1)^2

L = 3F - 1