Re: Common Multiple Value Question

*To*: mathgroup at smc.vnet.net*Subject*: [mg88226] Re: Common Multiple Value Question*From*: dh <dh at metrohm.ch>*Date*: Tue, 29 Apr 2008 06:47:25 -0400 (EDT)*References*: <fv42jv$5va$1@smc.vnet.net>

Hi Lea, I am not sure if I understand you correctly. You have two integer sequences, e.g. {a1,a2,a3} and {b1,b2,b3} and you want a sequence {c1,c2,c3} so that ai and bi divide ci, ci need not be the LCM and c1:c2:c3 == a1:a2:a3. This can be done by noting that a possible CM is simply the product. We then only have to deal with c1:c2:c3 == a1:a2:a3. This can be achieved by mutiplying a1 by b2 b3, a2 by b3 b1 .. This finally gives {a1,a2,a3} b1 b2 b3: In mathematica this may be written by: lista={a1,a2,a3}; listb={b1,b2,b3}; listc= lista Times@@ listb For rational number you must first define what a multiple of a number is. For every rational r1 besides zero there is another one r2 so that the product is a given r3. That is every number is a multiple of every number (besides zero). hope this the not right answer to the wrong question, Daniel Lea Rebanks wrote: > Dear All, > > I am trying to solve this Common Multiple Value Question based on 2 > sequences of values... > This is both a Math question & also a question on how to apply this problem > to Mathematica. > However, please note I am trying to find BOTH Common Multiple Value, of each > element in both lists AND maintaining the same proportion in the first list > of elements. > > I will pose this problem in 2 ways. > (i) Fractional sequence of values. > (ii) Numerical sequence of values. > Actually they are the same question, but I would like to be able to solve > this in both the fractional & numerical form. > (... And also solving this without using LCM - Lowest Common Multiple > function .... as this is part of a bigger calculation & I will be using > algebra.) > > (i) Fractional sequence of values - 1 pair. 3 elements in each. > list1={2/3,2/3+1,2/3+2} > list2={3/4,3/4+1,3/4+2} > {2/3,5/3,8/3} > {3/4,7/4,11/4} > > (ii) Numerical sequence of values - 1 pair. 3 elements in each. > list3={2,5,8} > list4={3,7,11} > {2,5,8} > {3,7,11} > > ( Apologies if this question is so simple... I have been trying to figure > this out for a number of days ...) > > Many thanks for any help or advice. Greatly appreciated. > Best regards ... Lea Rebanks. > >