
Re: List manipulation question  2013
Posted:
Jan 30, 2013 10:05 PM


list1 = Table[24 + (15*n1), {n1, 2000}];
Take[Select[list1/24, IntegerQ], 9]
{6, 11, 16, 21, 26, 31, 36, 41, 46}
Take[Select[list1, IntegerQ[#/24] &], 9]
{144, 264, 384, 504, 624, 744, 864, 984, 1104}
Apparently you want list2 to consist of the members of list1 that are integer multiples of 24 rather than the integers in list1/24.
list2 = Select[list1, IntegerQ[#/24] &];
list3 = n2 /. Solve[ {12 + (3*n2) == #, Element[n2, Integers]}, n2][[1]] & /@ list2;
Length[list2] == Length[list3]
True
Bob Hanlon
Consequently, there is an integer n2 for every element of list2.
On Wed, Jan 30, 2013 at 4:26 AM, Lea Rebanks <lrebanks@gmail.com> wrote: > > Dear All, > > I have the follow problem with the combination of a few lists. > I shall outline the problem or what I am trying to do. > I know that to solve this problem requires a number of processes, but I am > not sure how to setup the coding to achieve my desired result. > Please could someone show me the coding for this problem. > > The problem: > 1  I have a list1 created from 24+(15*n1) All n1 are integers > 1,2,3,4.....to a large number of integers. > 2  I want to divide the list1 by 24 and create another list2 with only the > integer results in list2. > Table[24 + 15*i, {i, 100}]/24 ... so that integer results in list2 > = { 144, 264, 384, 504, 624, 744, 864, 984, 1104 } > 3  I have another list3 created from 12+(3*n2) All n2 are integers > 1,2,3,4.....to a large number of integers. > 4  With list3 I want to find : > (i) The integer number of n2 that either equates 12+(3*n2) = 144 or > the next FIRST available number in list2 that meets this equality. > then also (ii)The integer number(s) of n2 that equates 12+(3*n2) = 24+(15*n1) after the first equal value for say 6 repetitions. > then also (iii)The integer number(s) of n1 that equates 12+(3*n2) = > 24+(15*n1) after the first equal value for say 6 repetitions. > > The above example is quite simple, but I am hoping to setup the coding to > work with other integer values instead of 3 & 15 which will present more of > a challenge. > > Any help or advice greatly appreciated. > Best regards, > Lea... > >

