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Re: find two numbers a,b, such that a+b=5432 & LCM[a,b]=223020
*To*: mathgroup at smc.vnet.net
*Subject*: [mg122159] Re: find two numbers a,b, such that a+b=5432 & LCM[a,b]=223020
*From*: "Harvey P. Dale" <hpd1 at nyu.edu>
*Date*: Mon, 17 Oct 2011 08:09:28 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <j6e693$kef$1@smc.vnet.net> <201110050801.EAA07025@smc.vnet.net> <360851F3-7690-47A6-BC84-E9521100E6F4@mimuw.edu.pl> <A0A54467-5287-4372-9FA6-F0FD865B77D2@mimuw.edu.pl> <84B4CB1D-C8A7-4A5F-8303-9FEC113F7D16@mimuw.edu.pl> <1318773921.12784.YahooMailNeo@web43139.mail.sp1.yahoo.com> <201110162045.QAA19701@smc.vnet.net>
Here's a simple brute-force method:
Select[Range[5431],LCM[#,5432-#]==223020&]
Best,
Harvey
-----Original Message-----
From: Andrzej Kozlowski [mailto:akoz at mimuw.edu.pl]
Sent: Sunday, October 16, 2011 4:46 PM
To: mathgroup at smc.vnet.net
Subject: [mg122159] Re: find two numbers a,b, such that a+b=5432 & LCM[a,b]=223020
Well, I don't think Reduce can solve it "directly" but here is an easy
method that makes use of Reduce and a tiny bit of elementary number
theory to solve it.
Let g denote the GCD of a and b. Then we know that a b = 223020 g
(from the well known relationship between GCD and LCM). Hence we need to
solve the equation Reduce[{a + b == 5432, a b == 223020 g}, {a,
b}, Integers] for all possible GCD candidates of a and b. Since the GCD
of a and b is always a divisor of their sum a+b == 5432 we can
simply test all possible divisors (there are only 16).
ls = Divisors[5432]
{1,2,4,7,8,14,28,56,97,194,388,679,776,1358,2716,5432}
(Reduce[{a + b == 5432, a*b == 223020*#1}, {a, b}, Integers] & )
/@ ls
{False, False, False, False, False, False,
(a == 1652 && b == 3780) || (a == 3780 && b == 1652),
False,
False, False, False, False, False, False, False, False}
Thus we see that the only possibility is that one of the numbers is 1562
and the other 3780.
Andrzej Kozlowski
On 16 Oct 2011, at 16:05, dimitris anagnostou wrote:
> Hello Mr Kozlowski .
>
> This is from a post of mine to MathGroup that it hasn't appeared yet.
>
> This is taken from the recent book (2010): " Mathematica: A
Problem-Centered Approach" by R. Hazrat.
>
> "The sum of two positive integers is 5432 and their least common
multiple is
> 223020. Find the numbers."
>
> A solution:
>
> Do[If[LCM[i, 5432 - i] == 223020, Print[i, " ", 5432 - i]], {i,
1, 2718}]
> 1652 3780
>
> I wonder if we can solve the system of equations:
>
> a+b==5432&&LCM[a,b]==223020
>
> using codes that contain built in functions like Reduce.
>
> I guess this is not a trivial one because the so much powerful Reduce
itself fails
>
> In[1]:= Reduce[{a + b == 5432, LCM[a, b] == 223020}, {a, b}, =
Integers]
>
> During evaluation of In[1]:= Reduce::nsmet:This system cannot be
solved with the methods available to Reduce. >>
>
> Out[1]= Reduce[{a + b == 5432, LCM[a, b] == 223020}, {a, b}, =
Integers]
>
> I would really appreciate your ideas.
>
> Best Regards
>
> Dimitris Anagnostou
>
> From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
> To: dimitris anagnostou <dimmechan at yahoo.com>
> Cc: Bobby Treat <drBob at bigfoot.com>; Dr. Wolfgang Hintze
<weh at snafu.de>; "mathgroup at smc.vnet.net Steve" <mathgroup at smc.vnet.net>;
Peter Pein <petsie at dordos.net>
> Sent: Thursday, October 6, 2011 9:48 PM
> Subject: Re: Re: simplification
>
>
> On 6 Oct 2011, at 20:28, Andrzej Kozlowski wrote:
>
> >
> > On 6 Oct 2011, at 16:57, Andrzej Kozlowski wrote:
> >
> >>
> >> On 5 Oct 2011, at 10:01, Peter Pein wrote:
> >>
> >>> Am 04.10.2011 07:40, schrieb dimitris:
> >>>> Hello.
> >>>>
> >>>> Let
> >>>>
> >>>> o1 = 1 + Sqrt[15 + 2*Sqrt[35] + 2*Sqrt[6*(6 + Sqrt[35])]];
> >>>> o2 = 1 + Sqrt[3] + Sqrt[5] + Sqrt[7];
> >>>>
> >>>> o1 is equal to o2.
> >>>>
> >>>> o1 == o2 // FullSimplify
> >>>> True
> >>>>
> >>>> The question is how to make Mathematica to simplify o1 to o2.
> >>>>
> >>>> Thanks
> >>>> Dimitris
> >>>>
> >>>
> >>> With a lot of luck:
> >>>
> >>> In[1]:= o1 = 1 + Sqrt[15 + 2*Sqrt[35] + 2*Sqrt[6*(6 +
Sqrt[35])]];
> >>> ext = Block[{x, poly = RootReduce[o1][[1]]},
> >>> Sqrt[Cases[Union @@ Divisors[Abs[CoefficientList[poly[x], x]]],
> >>> 1 | _?PrimeQ, 1]]]
> >>> o2 = ((Rest[#1] / First[#1]) . ext & )[
> >>> FindIntegerNullVector[Prepend[ext, -o1]]]
> >>>
> >>> Out[3]= {1, Sqrt[2], Sqrt[3], Sqrt[5], Sqrt[7], Sqrt[19],
Sqrt[31]}
> >>>
> >>> Out[4]= 1 + Sqrt[3] + Sqrt[5] + Sqrt[7]
> >>>
> >>> :-)
> >>>
> >>
> >> Neat, but from the Mathematical point of view the question was
posed "the wrong way round" in that o1 is mathematically "simpler" than
1 + Sqrt[3] + Sqrt[5] + Sqrt[7], since it is already expressed in terms
of its minimal polynomial. Hence this is the "natural" or "easy" way to
go:
> >>
> >> ToRadicals[RootReduce[1 + Sqrt[3] + Sqrt[5] + Sqrt[7]]]
> >>
> >> 1 + Sqrt[15 + 2*Sqrt[35] + 2*Sqrt[6*(6 + Sqrt[35])]]
> >>
> >> in other words, the algebraic "simplification" in this case is
exactly the opposite of, what might be called, the visual one.
> >>
> >>
> >> There is no natural or unique way to "decompose" algebraic numbers
that are already reduced into sums etc, of "simpler" summands or factors
etc. Of course, if we already know an integer basis for an algebraic
number field containing an algebraic number, than there are ways of
expressing it in terms of this basis - and this method is an example.
> >>
> >> Andrzej Kozlowski
> >>
> >>
> >>
> >>
> >
> > Let me correct myself here since I noticed that I did not write what
I really meant ;-)
> >
> > What I mean was that the following is the simplest (algebraically)
form of this algebraic number:
> >
> > In[80]:= RootReduce[1 + Sqrt[3] + Sqrt[5] + Sqrt[7]]
> >
> > Out[80]= Root[
> > 1024 + 3584 #1 + 640 #1^2 - 1984 #1^3 - 48 #1^4 + 304 #1^5 -
> > 32 #1^6 - 8 #1^7 + #1^8 &, 8]
> >
> > (the same as above, but, of course, without the ToRadicals).
Mathematically equivalent statement is:
> >
> > MinimalPolynomial[RootReduce[1+Sqrt[3]+Sqrt[5]+Sqrt[7]],x]
> > 1024+3584 x+640 x^2-1984 x^3-48 x^4+304 x^5-32 x^6-8 x^7+x^8
> >
> > The only simplification one can really make algorithmically with
algebraic numbers is what RootReduce does - essentially finding the
minimal polynomial. Mathematica also has an algorithm for converting
some root objects to radicals but this is usually does not give the
visually simplest form. There is no algorithm that will discover the
simples such form in general (indeed, most algebraic numbers can't be
expressed in radicals at all). On the other hand, one can find the
minimal polynomial of any algebraic number: e.g.
> >
> > In[82]:= MinimalPolynomial[1 + Sqrt[3] + Sqrt[5] + Sqrt[7], x]
> > Out[82]= 1024 + 3584*x + 640*x^2 - 1984*x^3 - 48*x^4 + 304*x^5 -
32*x^6 - 8*x^7 + x^8
> >
> > In[83]:=
MinimalPolynomial[1+Sqrt[15+2*Sqrt[35]+2*Sqrt[6*(6+Sqrt[35])]],x]
> > Out[83]= 1024+3584 x+640 x^2-1984 x^3-48 x^4+304 x^5-32 x^6-8
x^7+x^8
> >
> > You get the same answer which is how Mathematica knowns that these
numbers are really equal - it does not attempt to transform one into the
other. When I wrote that
> > 1+Sqrt[15+2*Sqrt[35]+2*Sqrt[6*(6+Sqrt[35])] was "algebraically
simpler" than 1+Sqrt[3]+Sqrt[5]+Sqrt[7]] I meant only that the former is
Mathematica's radical representation of Root[1024 + 3584 #1 + 640 #1^2 -
1984 #1^3 - 48 #1^4 + 304 #1^5 - 32 #1^6 - 8 #1^7 + #1^8 &, 8] - which
is indeed the simplest way (algebraically) to express this algebraic
number, which, of course, has infinitely many other representations in
terms of radicals (which Mathematica will not be able, in general, to
convert into one another but will always be able to find their minimal
polynomial).
> >
> > This is not a limitation of Mathematica - it is a limitation of the
radical representation of algebraic numbers.
> >
> > Andrzej Kozlowski
> >
> >
> >
>
>
> Even the above was still not quite accurate. MinimalPolynomial of an
algebraic number, still does not determine it - obviously
>
> MinimalPolynomial[
> Root[1024 + 3584 #1 + 640 #1^2 - 1984 #1^3 - 48 #1^4 + 304 #1^5 -
> 32 #1^6 - 8 #1^7 + #1^8 &, 8], x]
>
> 024 + 3584 x + 640 x^2 - 1984 x^3 - 48 x^4 + 304 x^5 -
> 32 x^6 - 8 x^7 + x^8
>
> MinimalPolynomial[Root[1024+3584 #1+640 #1^2-1984 #1^3-48 #1^4+304
#1^5-32 #1^6-8 #1^7+#1^8&,7],x]
> 1024+3584 x+640 x^2-1984 x^3-48 x^4+304 x^5-32 x^6-8 x^7+x^8
>
> but of course these roots themselves are not the same. However, once
you know the minimal polynomial of an algebraic number you only need to
isolate it from the other roots to determine it uniquely - which is
something that Mathematica does automatically when it numbers the roots.
So, once you determine the minimal polynomials of two algebraic numbers
we know that if the polynomials are different so are the numbers, if
they are the same we use root isolation to distinguish or identify them.
In any case, the point it that no transformations are performed on
radicals when doing this sort of thing.
>
>
>
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