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  • Apr 12, 2012, 11:24 AM
    shaun10
    Magic square
    How to construct a magic square by multiplying each of the numbers
    In the grid by 3
  • Apr 12, 2012, 12:06 PM
    ebaines
    I assume what you mean is a 3x3 grid where each square has a unique number and the numbers in all rows, columns, and the two diagonals multiply to the same constant. There is no requirement that the numbers must be consecutive.

    I found this article: Wolfram|Alpha: Math Reference: multiplication magic square which indicates that the lowest possible conrtant that all the rows, columns and diagonals can multiply to is 216. Consider the prime factors of 216:

    216 = 2 x 2 x 2 x 3 x 3 x 3

    So each square must have a number that is some multiple of 2's and 3's, or a 1. But no square can be a multiple of three 2's, or three 3's, because it would be impossible to have unique entries in the squares of the two or 3 intersecting rows & columns. So that leaves the following as possible numbers to use:

    1
    2
    3
    2 x 2 = 4
    2 x 3 = 6
    3 x 3 = 9
    2 x 2 x 3 = 12
    2 x 3 x 3 = 18
    2 x 2 x 3 x 3 = 36

    That's nine possible entries for the 9 squares of the grid. Now it's just a matter of playing with these numbers long enough. But note that the 36 can have only two possible combinations of other numbers to get to 216: either 1 & 6 or 2 & 3. Consequently 36 must be in a middle edge square, because numbers that are in a corner square must work with three sets of other numbers (one set each for the intersecting column, row and diagonal) and the center square number must work with 4 other sets of numbers (row, column, and two diagonals). Similarly 1 must be on a middle edge because the only two pairs of numbers that works with it are 12 & 18 and 6 & 36.

    OK, these are enough hints - can you complete it now?

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