19 votes
Accepted

Modify a magic square

How about: Check: An answer which is not of this form is:
elias's user avatar
  • 9,592
16 votes

Modify a magic square

My Shot: Reasoning. Second try. And maybe the simplest function
Marius's user avatar
  • 17.9k
15 votes

I'm trying to create a magic square

You need to:
JMP's user avatar
  • 35.6k
13 votes
Accepted

The magic square with a hole

Here are a few from this site. Any magic square with integers 0-15 will work, considering the 0 as a hole. I've chosen four with the 0 in the inner four cells just to be sure it meets your criteria. ...
Roland's user avatar
  • 9,733
13 votes
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Are there any sets of 9 numbers that can form two essentially distinct magic squares?

Note that any two magic squares with the same numbers must have the same row/column sum. The middle number must be 1/3 of the sum (because the sum of the four lines passing through the center is the ...
f'''s user avatar
  • 33.6k
13 votes

Number of magic squares with magic constant 0?

Take the example square below: -3 2 1 4 0 -4 -1 -2 3 To generate a new square, simply multiply each element of this square by any positive integer. As there ...
frodoskywalker's user avatar
12 votes
Accepted

Can you fill $3 \times 3$ magic square?

There are two ways to do this: the algebraic way and the 'clever' way. The algebraic way The clever way
Deusovi's user avatar
  • 145k
10 votes
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Not-Quite-Sufficiently-Advanced-Technology Square

I was able to find one of the solutions using "paper and pencil". I stopped searching for more solutions after that. It is certainly very very time consuming. In my explanation I name rows as A,B,C,...
ppgdev's user avatar
  • 1,305
10 votes
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Create a magic square of 4-digit numbers

Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total First of all, construct four single digit magic squares... Then concatenate ...
hexomino's user avatar
  • 133k
9 votes
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Magic square 4x4 that sum to 38

Here's how I would go about it: Firstly, we want to split the numbers in two groups, each having 8 numbers with a total of 76. We are going to use one of the groups for the diagonals, and the other ...
Bass's user avatar
  • 75.8k
9 votes
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A riddle that has been killing me the whole day

The answer is: The formula is: Examples:
nishuba's user avatar
  • 477
8 votes
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No ordinary magic square part 2. How many solutions are there?

According to the simple-minded 7-line Python program I just wrote, there are exactly There is at least a 25% chance that I didn't make any stupid blunders. Here's the code: ...
Gareth McCaughan's user avatar
8 votes
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Perfect magic 4x4 square

Solution: Notes:
loopy walt's user avatar
  • 19.1k
8 votes
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Four 3x3 semimagic squares in a 5x5 grid

user1502040's user avatar
7 votes

No ordinary magic square part 2. How many solutions are there?

My Python code to iterate the solutions and their respective $8$ sums: Counting them: Some more information:
Jonathan Allan's user avatar
7 votes

Unsolved Mysteries: Magic Square of Squares

Copied entirely from http://mathpages.com/home/kmath417.htm This is not a solution; it is just a MathJAXed version of the link provided. Magic Square of Squares It's an open question whether there ...
ghosts_in_the_code's user avatar
7 votes

Magic square using consecutive odd numbers -5 through 11

Take the standard $3 \times 3$ square, double each entry and subtract $7$. That is the linear transformation that takes $1$ to $-5$ and $9$ to $11$ $$\begin {array}{c c c} 9&-5&5\\-1&3&...
Ross Millikan's user avatar
7 votes
Accepted

Put numbers to a star-shaped puzzle

This does it: There may be other solutions beyond symmetry ...now to write some code new code for all valid solutions up to symmetry previous slow, non-symmetry "naive" search
Jonathan Allan's user avatar
7 votes
Accepted

How big can a witchcraft square be?

As @2012rcampion said, this is only possible when $n$ is even: The sum of all row and column sums of a witchcraft square is the sum of $2n$ consecutive natural numbers, and if $n$ is odd, this sum ...
Anon's user avatar
  • 2,684
7 votes
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Magic-preserving Permutations on a 4x4 Magic Square

Jaap Scherphuis's user avatar
7 votes
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It cannot be done. I think

Using only basic operators (+ - * / ^) There are 16 solutions that, like yours, are missing the center column: There are 64 boards total that are missing any one row, column, or diagonal. Here is the ...
DenverCoder1's user avatar
  • 10.4k
7 votes

Place the numbers 1-7 in the squares so that each row and columns adds up to the same total

The same 56 solutions but with logical reasoning instead of bruteforcing:
trolley813's user avatar
  • 11.3k
7 votes

How to solve 3x3 Magic Squares with negative values when only 2 values are given?

Turns out you don't have enough information -- but you could put anything you wanted in any other cell! As shown by Joe Z in this answer, all 3×3 magic squares can be expressed as: $$\begin{array}{|c|...
Deusovi's user avatar
  • 145k
6 votes
Accepted

An Antimagic Square

For any such magic square the sum of all rows, columns and diagonals: $$\sum_{x=0}^7{x}=28$$ Now each number is counted at least twice, once for the row and once for the column. Furthermore each ...
The Dark Truth's user avatar
6 votes

An Antimagic Square

The unique solutions (ignoring rotations and flips) are: Found through the method of full exhaustion.
dmg's user avatar
  • 5,294
6 votes
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What type of magic square is this?

The given square $X$ results from the following classical magic square (which arranges the number $1,2,\ldots,25$ in a $5\times5$ square so that every row, every column, and each of the two main ...
Gamow's user avatar
  • 45.4k
6 votes

Magic square with equal sums on rows, columns and diagonals

The answer is: Right So, here's an image Here's a description of the working:
Sid's user avatar
  • 15.1k
6 votes

Magic square 4x4 that sum to 38

One thing to consider is that there are only 4 odd numbers in the puzzle: 7, 7, 11, 13. Which means every row/column/diagonal that has one of those four numbers must have exactly two of them (you can'...
Adam S.'s user avatar
  • 406
6 votes
Accepted

Create a 3x3 Magic Square that uses integers from -10 to -2

Excited Raichu's user avatar

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