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Re: Common Multiple Value Question

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  • Subject: [mg88226] Re: Common Multiple Value Question
  • From: dh <dh at>
  • Date: Tue, 29 Apr 2008 06:47:25 -0400 (EDT)
  • References: <fv42jv$5va$>

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.



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