47

No such rectangle exists. Suppose you have a rectangle with $m$ rows and $n$ columns. If every row adds up to some magic value $M$, then the number obtained by adding together every cell in the rectangle must be $m \times M$. Likewise, if every column adds up to $M$, then the value obtained by adding together every cell in the rectangle must also equal $n \...


31

2 10 3 6 5 4 7 0 8 This is the only one, not counting reflections and rotations. Proof: As you found, the sum of all the numbers is three times the sum of each row (or column), so each row must add up to 15. There are only two ways to make 15 with 0 and two numbers: 0+7+8 and 0+5+10. Similarly, 10 can only be used in 2+3+10 and 0+5+10. Each corner ...


22

In general, to know whether this sort of problem is solvable (given a list of 9 numbers, form a magic square out of them), you need to look for three arithmetic sequences that are themselves arithmetically sequenced — namely, three sequences of the form: \begin{matrix} a & a+c & a+2c \\ a+b & a+b+c & a+b+2c \\ a+2b & a+2b+c & a+...


22

A possible solution is: 10 8 3 21 12 15 14 2 1 7 30 24 42 6 4 5 Strategy $$5040=2^4 \times 3^2 \times 5 \times 7$$ First I decided where to put the multiples of $7$ and $5$. Then I multiplied proper exponents of $2$ and $3$ to each cell. I started with: 5 1 1 7 1 5 7 1 1 7 5 1 7 1 1 5 This formation ensures that the number of ...


19

This is just a standard Lo Shu Square with exponents and multiplication instead of addition. But because the multiplication of exponents with the same base is analogous to the addition of their exponents, the solution here is just the same as $2^\left(L-1\right)$, where $L$ is the standard Lo Shu Square. As an expanded explanation, your numbers are better ...


19

How about: Check: An answer which is not of this form is:


17

My Shot: Reasoning. Second try. And maybe the simplest function


16

It turns out that the magic squares form a vector space. You can add them (by adding each corresponding entry), multiply them by scalars (by multiplying every entry by the same number), and invert them (take the negative of each entry) around a zero element (the magic square where everything is zero), and they will still stay magic squares. And this vector ...


16

Here's the solution I created: $$\begin{array}{ccc|ccc|ccc} 70 & 63 & 68 & 7 & 0 & 5 & 52 & 45 & 50\\ 65 & 67 & 69 & 2 & 4 & 6 & 47 & 49 & 51\\ 66 & 71 & 64 & 3 & 8 & 1 & 48 & 53 & 46\\\hline 25 & 18 & 23 & 43 & 36 & 41 & 61 & 54 & ...


16

Every normal magic square of order three is obtained from the Lo Shu by rotation or reflection. From Wikipedia: Lo Shu Square Since rotation and reflection cannot move the 8 out of a corner, such a square is impossible.


16

With a brute force program solver written in C#, I found a solution with sum 366: 3 11 5 | 19 37 17 13 | 67 7 31 23 | 61 ---------+--- 29 47 59 41 | 43 I let the program run until the top left corner was 61, so I'm pretty sure that there are no better solutions, but feel free to look for yourself: Source code at PasteBin (you might want to ...


15

The squares you describe are related to Latin squares. A Latin square of order $n$ is an $n\times n$ square grid, filled with the numbers $1,\ldots,n$, such that each number appears once in every row and in every column. A Latin square is called doubly diagonal if each number also appears once on each diagonals. A doubly diagonal Latin square satisfies your ...


15

There is a general, very simple, algorithm for generating any magic square which has an odd number of rows/columns as follows: Start in the middle of the top row and enter 1. Move Up 1 and Right 1 wrapping both vertically and horizontally when you leave the grid *(see note below). If that square is empty enter the next number; if the square is not empty ...


15

Given your description, the filled-in numbers on the magic square look like this: $$\begin{array}{c|c|c} &1&\\\hline 7&&\\\hline &&9 \end{array}$$ Now, other answerers have pointed out that the middle number is 5, so the 9 should go on the bottom rather than the bottom-right, in which case the completed magic square looks like this: ...


15

You need to:


13

As Kendall mentions, it is impossible to move pieces out of a 3x3 grid in this manner. This is called the Lo Shu square, and is the only unique solution to the 3x3 magic grid puzzle. The reason for this is actually rather simple. If you look at the Lo Shu square, you'll notice that the even numbers always appear in corners. To demonstrate why this would be, ...


13

In general, any $n\times n$ magic square of range [1, $n^2$] with odd number $n$ can be solved using the following algorithm: Start at the middle grid in the bottom row. This is your 1. Move downwards and to the right by one grid. If this move results in a position outside the square, wrap around to the beginning of the row (or column). If 2 cannot be ...


13

The mathematician's task is Proof:


13

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 sum of all the numbers plus three times the center), so it is the same in both squares. Subtract this number from all the numbers, so that the middle cell and ...


13

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 are an infinite number of positive integers, there are an infinite number of possible squares.


12

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. The rows, columns, and diagonals of each add up to 30.


10

The answer to the first one is: The answer to the second one is: The answer to the last one is: To solve these, the easiest way is to:


10

The easiest way is just to take all 23's and... So the solution is:


10

Another solution (with diagonals as bonus): 10 4 6 21 18 7 20 2 28 12 3 5 1 15 14 24 Things that multiply to 5040: each of the four rows each of the four columns each of two diagonals four center cells four corner cells two middle cells of the top row two middle cells of bottom row two middle cells of the rightmost column ...


10

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,D,E - columns as 1, 2, 3, 4, 5. My search strategy is based on an observation that As you can see this gave me only 4 possible values for $B1$ Page 2 ...


9

This is a two-part answer. First, it establishes a non-trivial upper bound of 5. Then a solution is given that proves 5 is also the lower bound. There is only one 3x3 magic square, up to symmetry: 294 753 618 So in order to have seven magic squares in a Sudoku, we require seven 5s which aren't against an edge, and that requires two diagonally opposite ...


9

Given an $n\times n$ magic square, write $M_n$ for its magic constant. There are $n$ rows, each of which has sum $M_n$, so the sum of all the entries in the square is $n \cdot M_n$. Each whole number between $1$ and $n^2$ appears once, so $$ \begin{gather} n\,M_n=1+2+\ldots+n^2=\frac{n^2(n^2+1)}{2},\\ M_n = \frac{n(n^2+1)}{2}. \end{gather} $$ In particular, ...


9

I have found a minimal solution. First of all, the puzzle asks for $31$ distinct non-negative numbers. If we take $0$ as a non-negative as well, the $31$ smallest integers are $0 - 30$ and they sum up to $(30*31)/2 = 465$. Now, it should be possible to divide this grand sum through both $4$ and $9$ and that is not possible with $465$. We need a multiple of $...


9

According to the article, the Magic Hexagon is the only one that exists (aside from n=1). There wouldn't be a simpler method for generating one, because no other hexagons of order 2 or >3 would have a solution. Quoting from the last paragraph: Trigg showed that the magic constant for an order n hexagon would be (9(n^4-2n^3+2n^2-n)+2)/(2(2n-1)), the ...


9

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 for the edges. To be able to construct the diagonals, at least one of the groups needs to be further splittable into two equal parts of 4 numbers of total 38. ...


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