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## Hot answers tagged magic-square

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 How about: Check: An answer which is not of this form is: 17 My Shot: Reasoning. Second try. And maybe the simplest function 16 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 ... 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: $$\begin{array}{ccc|ccc|ccc} 70 & 63 & 68 & 7 & 0 & 5 & 52 & 45 & 50\\ 65 & 67 & 69 & 2 & 4 & 6 & 47 & 49 & 51\\ 66 & 71 & 64 & 3 & 8 & 1 & 48 & 53 & 46\\\hline 25 & 18 & 23 & 43 & 36 & 41 & 61 & 54 & ... 16 Every normal magic square of order three is obtained from the Lo Shu by rotation or reflection. From Wikipedia: Lo Shu Square Since rotation and reflection cannot move the 8 out of a corner, such a square is impossible. 16 With a brute force program solver written in C#, I found a solution with sum 366: 3 11 5 | 19 37 17 13 | 67 7 31 23 | 61 ---------+--- 29 47 59 41 | 43 I let the program run until the top left corner was 61, so I'm pretty sure that there are no better solutions, but feel free to look for yourself: Source code at PasteBin (you might want to ... 15 The squares you describe are related to Latin squares. A Latin square of order n is an n\times n square grid, filled with the numbers 1,\ldots,n, such that each number appears once in every row and in every column. A Latin square is called doubly diagonal if each number also appears once on each diagonals. A doubly diagonal Latin square satisfies your ... 15 Given your description, the filled-in numbers on the magic square look like this:$$\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 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 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. 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. 10 According to Kendall and others, the puzzle isn't solvable for integers. This can be proved using the Lo Shu Square. But the problem statement says "A magic square (of order 3) is a 3x3 matrix consisting of distinct numbers from 1 to 9", and doesn't explicitly specify integers, a solution with real numbers could be obtained. Here's one: +---+---+---+ |4.5|... 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 ... 10 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 them to get a 4-digit magic square! 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 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, ...

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