anonymous
  • anonymous
HELP! URGENT! Need simple formula for "greatest common divisor" for only 2 numbers (integers)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Ok ! For example 32 & 84 First write : 84=X^y 32=X^y Means it : 32 = 2^5 84=2^4 . 3 Now get the smallest : It s 2^4 Then greatest common divisor = 2^4 = 16 Understand ?!:) :D It has another way too !
anonymous
  • anonymous
@E.ali, thank you) could you please show me other ways? I'm currently working on programming, so it would be great if there were a particular formula.
amistre64
  • amistre64
i wrote up euclids algorithm once

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amistre64
  • amistre64
L/S = Q.xxxx -Q ------- 0.xxxx times S -------- R L = S, S=R, repeat have to recall how i determined a break in the loop tho
amistre64
  • amistre64
if R = 0, then the gcd is the previous R
anonymous
  • anonymous
Now look : |dw:1062636136311:dw|
anonymous
  • anonymous
@amistre64 could you please be more specific? I feel quite confused, however, your method seems to be more suitable for my condition
anonymous
  • anonymous
@SerikMB : It s an interesting way !
anonymous
  • anonymous
Now we have 84/32 Answer to it!
anonymous
  • anonymous
2.625
anonymous
  • anonymous
what's next?
amistre64
  • amistre64
im hammering out some javascripting at the moment ... debugging
anonymous
  • anonymous
@E.ali actually i think the GCD for 84 and 32 is 4
anonymous
  • anonymous
OK ! Thank you ! But we want just it :|dw:1062637723502:dw|
anonymous
  • anonymous
Understand ?!
anonymous
  • anonymous
sorry, but i still can't get in ur method with boxes
anonymous
  • anonymous
OK friend ! It s is that 84/32 we have :|dw:1062637919084:dw|
amistre64
  • amistre64
Number 1:
Number 2:

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amistre64
  • amistre64
ive got to run to class for an hour :)
anonymous
  • anonymous
@amistre64 cool job BTW )) thank u!))
anonymous
  • anonymous
@E.ali what's next, we got 84, 32, 20 and 2
anonymous
  • anonymous
wow that is cool
amistre64
  • amistre64
i had to fix a few slight bugs. As is it has to go thru at least 1 cycle. I added some lines to fix it: if A == B, then gcd = A (the gcd of a number with itself is itself) if g==0 in the end, then the gcd is the smaller of the numbers. 12 = 6(2) + 0 has only one cycle
amistre64
  • amistre64
Number 1:
Number 2:
The GCD is:
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amistre64
  • amistre64
lol, once last bug to squash ... the gcd(0,n) = n ; and gcd(0,0)=0, and that should do it for my concept :)
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