xeda(a,b) © a, b numbers= parameters Prgm Local c,d,g gcd(a,b)->c newList(c)->q 1->g iPart(sqr(c))->d While d>0 © While d>1={}(empty list output) if a=2 and b=4 for ex. While mod(c/d,1)>0 © We need add a condition in 68k basic. d-1->d EndWhile d->q[g] g+1->g If c/d ≠ d Then © If block: for the sake of number c/d->q[g] © duplicating in the q list. EndIf d-1->d © Without this, infinite loop. EndWhile left(q,g-1)->q © q list without zeros. SortA q disp q EndPrgm

BTW2:
"I am not sure if factor() can be used on integers, but if it can, use it."
Yes. If the 2 numbers are a=20 and b=40 for example, the shortest way to calculate all the common divisors without programming is this:

factor(gcd(20,40)) = 2^2*5

So the divisors are: 2, 2^2, 2*5, 2^2*5.
I like this text formatting. ;)

If you convert List P to string, you can search in it with the inString command. If the result is not 0, it's a new common divisor.

common()
Prgm
Local pstr,qsi,n,nr
string(p)->pstr
dim(q)->qsi
0->nr
newList(qsi)->res
For n,1,qsi
If inString(pstr,string(q[n]))<>0 Then
nr+1->nr
q[n]->res[nr]
EndIf
EndFor
left(res,nr)->res
Disp res
EndPrgm

How can I insert here a formatted text?

Input A Input B gcd(A,B)→C ;All you have to do is find all the factors to this {C}→q 1→G iPart(sqrt(C))→D While D>1 While fPArt(C/D) ;I would have just used mod(), but I am not sure how the TI-89 handles mod 1. D-1→D EndWhile D→q[G] G+1→G C/D→q[G] G+1→G EndWhile SortA(q)

I've never had much practice with 89 BASIC, so some of the syntax might be wrong and I haven't tested the code yet, but hopefully it works. I am not sure if factor() can be used on integers, but if it can, use it.

In case you don't get why all you need to do is factor the GCD of the two numbers, think about it— the Greatest Common Divisor has all the mutual factors. If you do A/GCD(A,B) and B/GCD(A,B), they are by definition relatively prime (so they have no common factors). As a side note for the curious, the code I provided is not even close to polynomial time speed, but GCD itself is a polynomial time algorithm (and quite simple, too).

I like math >.>

For(A,1,*first#*
If *first#* is divisible by A
store A to list1
End
For(A,1,*second#*
If *second#* is divisible by A
store A to list2
End
For(A,1,*length of list1*
If list1(A) is in List 2
store # to list3
End
Clear lists1and2
list3 has your common divisibles

Sorry if that doesn't help really because I don't know the language

To accomplish this task I need some help,

One element of this program is taking the two numbers dividing them by a common integer then storing each number after division to a list

:1→N :1→G {0,0,0}→q This stores (0,0,0) to a the variable q, but also makes it a list {0,0,0}→r :Input A Input two numbers :Input B : :A/n→q[g] Division by common number, Q and R are lists with g being the position the divided number is being stored. :B/n→r[g] :

Now the really tricky part is here.

How do i cross check each of the lists to find a common number besides using SortA and SortD since they have limited use?

Ex. :List Q {5,6,7} :List R {6,8,9} : : : Using SortA or SortD would not work to cross check the lists even though there is a common number.

*SideNote I typed this all by hand so if the coding looks funky thats my bad*

Please comment to help me in TI-89 Format with somewhat of an explaintion of your coding.
-Thanks Keynah

(I will check this in 2-3 days , 6/8/2012)