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Ultramagic square

Posted on: February 11th, 2014 by
2

A 9x9 grid is filled with numbers from 1 to 81, each number occurring exactly once. The grid is said to be ultramagic if the product of numbers in row k is equal to the product of numbers in column k for each k. Can you construct an ultramagic grid?

- via AMS puzzle corner


2 Responses to Ultramagic square

  1. Dinesh had this to say about that:

    No such grid exists. This can be seen from the location of the primes in the grid. Let a prime p be located in position (i,j) for j \neq i. Then, the ith row product is divisible by p. The grid construction rule then stipulates that the j column is also divisible by p. This is only possible if a multiple of p (2p,3p...) is located on the jth column. Particularly, it is not possible for any prime above 81/2. So, the primes between 41 and 81 cannot lie on the off-diagonal. But there are 10 of them and only 9 diagonal locations.

    For a general n, the above argument yields a necessary condition \pi(n^2) - \pi(\lfloor \frac{n^2}{2} \rfloor) \leq n which rules out n \geq 11. Here \pi(.) is the prime counting function. It will be very surprising if any ultramagic squares exist at all.

    • Sid Hollander had this to say about that:

      Great logic. I question the posts' definition of ultramagic square in the puzzle. Is the requirement re. products a general one (where) or is it particular to this site only?


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