I have an idea, which is to put 2 in the middle and have the rest multiply up to an even number, but I can't seem to find an even number with that many factors.
Fill out a 3x3 square with 9 different positive integers such that the product of each row, column, and diagonal is equal to each other
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magic-square
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0Just for clarification, by 9 different positive integers you don't just mean $1,2,\dots,9$? – 2017-02-18
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0You can use any positive integers. – 2017-02-18
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0Nevermind, it's unsolvable if only 1 to 9 are allowed – 2017-02-18
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4If you write each of the 9 entries as a power of 2 with an unknown exponent, then it will work provided the 3x3 square with just the exponents is a magic square with all distinct entries. – 2017-02-18
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1please check: (http://puzzling.stackexchange.com/questions/3607/can-you-fill-a-3x3-grid-with-these-numbers-so-the-products-of-the-rows-and-colum) – 2017-02-18
1 Answers
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Hint : Take a magic $3\times 3$-square and replace every entry $x$ by $2^x$