Solving a 3 x 3 Magic SquareDate: 09/29/2005 at 19:59:57 From: Mick Subject: Magic Squares Place the numbers 1-9 in a 3 by 3 grid, one number per box, so that the vertical, horizontal, and diagonal sums are all the same. I have tried and tried to figure this out, but I can't seem to get it. I know that small numbers should be on the end because that is what our teacher said, and I'm pretty sure the middle number is 9 but I just haven't been able to find the answer. Date: 09/30/2005 at 01:19:44 From: Doctor Wilko Subject: Re: Magic Squares Hi Mick, Thanks for writing to Dr. Math! You could certainly use a "trial and error" approach and get a correct solution to the 3x3 magic square, but it sounds like you've tried that without luck, so let's try a more methodical approach. One way a 3x3 magic square can be constructed is by using a little simple arithmetic and algebraic reasoning. One could reason that the numbers 1-9 add up to 45, and since all these numbers are contained in three ROWS (exclusively), the sum of each ROW must be 15. I want to make sure you understand this before proceeding, so look at the little algebraic argument below that "proves" that each row should sum to 15: Since we don't know the order of the numbers 1-9 that correctly fill out the 3x3 magic square, substitute the variables a-i into the square. +---+---+---+ | a | b | c | +---+---+---+ | d | e | f | +---+---+---+ | g | h | i | +---+---+---+ We know that a + b + c + d + e + f + g + h + i = 45 grouping the letters by rows we get (a + b + c) + (d + e + f) + (g + h + i) = 45 But since the sum of each row should be equal (let's say the sum equals X), the rows can be rewritten as (a + b + c) = X (d + e + f) = X (g + h + i) = X and the equation above can then be rewritten as X + X + X = 45, or 3*X = 45, so X = 15 This shows that the three rows sum to 15. It then follows from the definition of a magic square that all the columns and diagonals will also sum to 15. OK, now we know that each row, column, and diagonal will sum to 15, so the challenge is to figure out HOW to arrange the numbers to make that happen. This may be done by a brute-force method as follows. Basically, through trial and error, I found all the ways that I could take three numbers from 1-9 and get a sum of 15: 1 + 5 + 9 = 15 1 + 6 + 8 = 15 2 + 4 + 9 = 15 2 + 5 + 8 = 15 2 + 6 + 7 = 15 3 + 4 + 8 = 15 3 + 5 + 7 = 15 4 + 5 + 6 = 15 Again, double check me here! Convince yourself that these are the only ways you can get a sum of 15 from three numbers chosen from 1-9. Now, from the equations above, note the following connections to the 3x3 magic square: 1, 3, 7, and 9 are each in TWO equations that sum to 15. The middle cells of each outside row/column are each in TWO equations (One row, one column). +---+---+---+ | - | x | - | +---+---+---+ | | | | | +---+---+---+ | | | | | +---+---+---+ (I'm willing to bet that 1, 3, 5, and 7 will each fill one of these middle cells on the outside rows/columns!) 2, 4, 6, and 8 are each in THREE equations that sum to 15. The corner cells are each in THREE equations (One row, one column, and one diagonal). +---+---+---+ | x | - | - | +---+---+---+ | | | \ | | +---+---+---+ | | | | \ | +---+---+---+ (From this, I'd be willing to bet that 2, 4, 6, and 8 are going to fill the corners of the magic square!) 5 is in FOUR equations that sum to 15. The middle cell is in FOUR equations (one row, one column, and two diagonals). +---+---+---+ | \ | | | / | +---+---+---+ | - | x | - | +---+---+---+ | / | | | \ | +---+---+---+ (From this it seems likely that 5 should go in the middle cell of the 3x3 magic square!) --------------------------------- Using the above observations, let's fill out the square! Looking at a blank 3x3 magic square, one can see that the middle cell should be in four equations (one row, one column, and two diagonals), and looking at the equations, one can see that 5 is the only number in four of the eight equations. Therefore, 5 can be put into the middle cell. +---+---+---+ | | | | +---+---+---+ | | 5 | | +---+---+---+ | | | | +---+---+---+ Now there are four corners which when filled, are each used in three equations. By putting any of the four numbers (2,6,4,8) in the upper-left corner (2 in this example), only one number can go in the bottom-right corner (8 in this example) to make the sum of that diagonal be 15. +---+---+---+ | 2 | | | +---+---+---+ | | 5 | | +---+---+---+ | | | 8 | +---+---+---+ Now, either of the other numbers (4 or 6 in this example) can go in the bottom left corner (6 in this example), leaving only one choice for the top right corner (4 in this example). +---+---+---+ | 2 | | 4 | +---+---+---+ | | 5 | | +---+---+---+ | 6 | | 8 | +---+---+---+ Last, the remaining four numbers (1, 3, 7, 9) can be placed in only one way to make the sums be 15. +---+---+---+ | 2 | 9 | 4 | +---+---+---+ | 7 | 5 | 3 | +---+---+---+ | 6 | 1 | 8 | +---+---+---+ This is one correct solution to the 3x3 magic square. There are seven other solutions. A 3x3 magic square contains eight possible solutions total; four rotations and four reflections. For more information on magic squares, you can visit the following links from our site: Magic Squares http://mathforum.org/alejandre/magic.square.html Magic Squares http://mathforum.org/library/drmath/view/67016.html Does this help? Please write back if you have further questions. - Doctor Wilko, The Math Forum http://mathforum.org/dr.math/ |
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