Replacement goes a long way. You may take all the 3s in the square below and turn them into 29s. Or some other square having large primes may be replaced with a square with smaller corresponding primes. Even primes within a particular square may be switched. 55440=2^4*3^2*5*7*11 so you can make a corresponding square with 2*3^2*5*7*11^4. No matter the starting square, this reduction leads to some sort of minimal representation, where the factorization of the product gives 2 with the largest exponent, 3 the next and so on. -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Michael Kleber Sent: Wednesday, September 21, 2005 2:32 PM To: math-fun Subject: Re: [math-fun] Multiplicative Magic Squares Just blathering off the top of my head: Ed pegg wrote:
Are multiplicative magic squares well studied? For example, using dots merely as spacers, the following has a multiplicative constant of
55440.
231 .40 ..3 ..2 ..6 ..1 132 .70 ..4 .21 .20 .33 .10 .66 ..7 .12
For any prime p, this must still be multiplicatively magic if you replace each entry with the largest power of p dividing it. And that must be additively magic if you take its log_p. For instance, for p=2, the above becomes 0 3 0 1 1 0 2 1 2 0 2 0 1 1 0 2 with additive constant 4, the power of 2 in 55440. Of course, this projection destroys distinctness of entries. For that you just need to find another one like the above such that the ordered pairs of corresponding entries (a,b) are all distinct, and then make a multiplicative one out of 2^a 3^b. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun