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[math-fun] Mulgic Squares
by Schroeppel, Richard 22 Sep '05

22 Sep '05

Minimality of K = 216 for 3x3 mulgic square: Proof sketch: K must be a cube, with at least 9 divisors. 1,8,27,64,125 have 1,4,4,7,4 divisors. Done. -----Original Message----- From: Schroeppel, Richard Sent: Thursday, September 22, 2005 10:07 AM To: 'math-fun' Cc: Schroeppel, Richard; 'rcs(a)cs.arizona.edu' Subject: RE: [math-fun] Multiplicative Magic Squares There's a standard proof that the center of a 3x3 magic square is K/3, where K is the row sum. (Add up the four lines through the center, subtract the whole square.) Of course this works for multiplication too, so the magic product is always the cube of the center. There's a collection of similar results for a 4x4 square, such as "the sum of the four corners is K" etc. Although the arguments offered so far are plausible, I think a bit more proof is in order before claiming minimality of the various squares. Maybe a program? There aren't that many potential products < 5040 with 16 divisors, and we can also require the exponents in the prime factorization form a descending staircase. Allowing 0 as an entry doesn't seem helpful. I suspect that allowing negatives will turn out to be equivalent to the positive case with duplications allowed (but not triplications & up). But I don't see the proof that minus signs can always be attached to a positive square in a consistent way. Rich rcs(a)cs.arizona.edu -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Christian Boyer Sent: Thursday, September 22, 2005 9:02 AM To: 'math-fun' Subject: RE: [math-fun] Multiplicative Magic Squares Of course replace "magic sum" by "magic product", twice in my previous message... -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr(a)mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : jeudi 22 septembre 2005 16:30 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3 _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Multiplicative Magic Squares
by Torgerson, Mark D 22 Sep '05

22 Sep '05

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(a)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(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Multiplicative Magic Squares
by Schroeppel, Richard 21 Sep '05

21 Sep '05

Start with the old 4x4 magic square construction 1 - - 4 - 15 14 - - 6 7 - merged with 12 - - 9 - 10 11 - 8 - - 5 13 - - 16 - 3 2 - Subtract 1 (to make the range 0-15) and then take binary bit planes. This gives the patterns --xx -x-x -xx- -xx- xx-- x-x- -xx- x--x xx-- x-x- x--x -xx- --xx -x-x x--x x--x and each row, column, and diagonal has two xs. Stacking the four patterns gives all 16 possible Z combinations. Assign the primes 2,3,5,7 to the four patterns. Multiply the pattern entries, getting all the divisors of 210 as the square entries. The magic product is 210^2 which is 44100. We can lower the product a little by using 4 in place of 7; the product is then 14400. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com]On Behalf Of ed pegg Sent: Wednesday, September 21, 2005 3:50 PM To: math-fun Subject: RE: [math-fun] Multiplicative Magic Squares If you just use powers of 2, the minimal square with distinct entries has multiplicative constant 1073741824 (2^30). 55440 is much smaller. Is it the smallest? --Ed Pegg Jr --- "Torgerson, Mark D" <mdtorge(a)sandia.gov> wrote: > 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. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] Multiplicative Magic Squares
by Torgerson, Mark D 21 Sep '05

21 Sep '05

Just looking and hoping my finger arithmetic is correct. In the 55440 square, I think you can turn 11s into 3s and the 7s into 5s giving 2^4*3^3*5^2=10800. -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of ed pegg Sent: Wednesday, September 21, 2005 3:50 PM To: math-fun Subject: RE: [math-fun] Multiplicative Magic Squares If you just use powers of 2, the minimal square with distinct entries has multiplicative constant 1073741824 (2^30). 55440 is much smaller. Is it the smallest? --Ed Pegg Jr --- "Torgerson, Mark D" <mdtorge(a)sandia.gov> wrote: > 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. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] 177 million tons of ice???
by Steve Rowley 21 Sep '05

21 Sep '05

>Date: Wed, 21 Sep 2005 13:46:23 -0700 (PDT) >From: Eugene Salamin <gene_salamin(a)yahoo.com> > >--- Steve Rowley <sgr(a)alum.mit.edu> wrote: >> Just put a magnet in your water bottle and you're ready to sell >> "ortho-water bottles", since the magnetic field will make the ortho >> polarization energetically favored. Slap a little sermon on the side >> about statistical thermodynamics and the magical incantation >> "exp(-uH/kT)", and it might even become trendy. "Refrigerate for >> higher polarization", that sort of thing. > >This is good marketing. It combines magnets with "structured water". >Now throw in homeopathy too. Homeopathy? Wow, that's even more twisted than I could think of... :-) >> You could even sell "after-market ortho-water bottle adapters" (aka >> magnets), to tape onto existing water bottles. > >No, no, no, disposibles are where the money is. Right, disposable magnets! Even better. (See? Don't ever let anybody tell you peer review doesn't work.) >> Strangely enough, spin-polarized hydrogen has occasionally been >> mentioned as a very dense energy storage mechanism. Its instability >> makes it more of a bomb than a gas tank, alas. > >This is electron spin, not nuclear spin. Hydrogen atoms with opposite >electron spin can bind to form a molecule, whereas atoms with the same >spin cannot. So by polarizing a gas of atoms, giving them all the same >spin, you keep them in a high energy state. I know; I was joking. A bit of Googling ressurects the fact that it was Tom Greytak and Dan Kleppner at MIT who were doing this back in the 80's: http://web.mit.edu/physics/greytak-kleppner/ -- Steve Rowley <sgr(a)alum.mit.edu> http://alum.mit.edu/www/sgr/ ICQ: 52-377-390

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Re: [math-fun] 177 million tons of ice???
by Henry Baker 21 Sep '05

21 Sep '05

Yes, I think that this is on purpose, since the original defn of a ton was used for the displacement of water in a ship. According to www.thefreedictionary.com/ton , a "ton" is also a unit of _volume_ in shipping -- either 35, 40 or 100 cubic feet. 35 cubic feet is approx one cubic meter. I recall from high school chemistry that a cubic foot of water is 64#, so 64*35 = 2240#, which is the definition of a "long ton". At 10:04 AM 9/20/2005, wrote: >Henry wrote: > ><< >177 million tons should be the weight of a small comet or a decent >size meteor (see <http://www.meteorcrater.com/eventsfun/exptheimp.htm>www.meteorcrater.com/eventsfun/exptheimp.htm) I'm >not sure that whether the meteor was iron or water would make much >difference in the resulting crater. >>> > >In discussions about this, a surprising factoid I learned is that >the volume of 1 ton of ice is almost exactly 1 cubic meter. >This is an easy rough estimate, but I'd never thought of doing it >before. At first blush, that a ton of ice should be a mere cubic meter >was amazing to me. (Either a ton is a lot smaller than it seems, >or a cubic meter is much bigger.) > >--Dan

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Re: [math-fun] 177 million tons of ice???
by Steve Rowley 21 Sep '05

21 Sep '05

Just put a magnet in your water bottle and you're ready to sell "ortho-water bottles", since the magnetic field will make the ortho polarization energetically favored. Slap a little sermon on the side about statistical thermodynamics and the magical incantation "exp(-uH/kT)", and it might even become trendy. "Refrigerate for higher polarization", that sort of thing. You could even sell "after-market ortho-water bottle adapters" (aka magnets), to tape onto existing water bottles. Maybe people will think they can only drink it while facing north. [Ok, that's enough flirting with The Dark Side for me this month. :-] >Date: Wed, 21 Sep 2005 13:58:09 -0400 >From: Tom Knight <tk(a)csail.mit.edu> > >I see a market opportunity. Think about the ad campaign for *pure* >ortho water. >You *do* want all of your water be straight, don't you! > >On Sep 21, 2005, at 1:53 PM, Eugene Salamin wrote: >> I found a great web site on water: http://www.lsbu.ac.uk/water/ . [...] >> In addition to the isotopic molecular species due to H, D, and the >> radioactive T, and also the stable O-16, O-17, and O-18, there is >> another class of distinct molecules that differ in the alignment of >> their nuclear spins. Considering only H2O, the H nuclei (protons) have >> spin 1/2. In ortho-water, the two spins are parallel, resulting in >> total nuclear spin 1, while in para-water, the spins are antiparallel, >> resulting in total nuclear spin 0. The equilibrium ratio is all para >> at 0 K, and 3:1 ortho:para at high temperatures (> 50 K). The >> equilibration time is about 1 hour in liquid water and several months >> in ice. It appears possible to separate ortho and para-water, but I >> can't say more because I don't have access to the journal articles from >> here. [...] >> Similar considerations apply to the hydrogen molecule H2. When >> hydrogen gas is liquified, it retains the 3:1 ratio. As ortho-hydrogen >> slowly converts to para-hydrogen at cryogenic temperatures, the energy >> released causes a substantial evaporization of the liquid. For this >> reason, a catalyst is used to quickly convert the hydrogen to para >> form. This can double the storage lifetime of the liquid. Strangely enough, spin-polarized hydrogen has occasionally been mentioned as a very dense energy storage mechanism. Its instability makes it more of a bomb than a gas tank, alas. -- Steve Rowley <sgr(a)alum.mit.edu> http://alum.mit.edu/www/sgr/ ICQ: 52-377-390

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RE: Antwort: [math-fun] Floor function question
by dasimov＠earthlink.net 21 Sep '05

21 Sep '05

Hartmut writes: << The real numbers are characterised as being the unique totally ordered field, s.t. for every number x there exists a natural number n (= 1+...+1) with n>x. The set of all n with n>x has a unique minimum m =: ceiling(x). floor(x) is then defined in the usual way. A total ordering is necessary (complex numbers), that every number is eventually surpassed by the naturals is necessary as well (surreal numbers). >> Wouldn't *any* subfield F of the reals R also be a totally-ordered field satisfying the same condition? --Dan

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[math-fun] Floor function question
by David Wilson 21 Sep '05

21 Sep '05

Starting with a field-based axiomatization of the real numbers, how do we show that floor(x) exists for every x? -------------------------------- - David Wilson

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Re: [math-fun] 177 million tons of ice???
by dasimov＠earthlink.net 20 Sep '05

20 Sep '05

Tom writes: << The molecular size and charge distribution is almost identical, so the crystals form without distinguishing between H2O and DH2O. It would be rare to find any significant number of molecules of D2O. DH2O ice does not sink, in any case. >> I'm confused. Does DH2O mean DHO (one deuterium, one hydrogen, one oxygen)? (If not, please define. If so, why the 2 ?) --Dan

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