Greatest Common Factor (GCF)

Date: 04/13/99 at 10:38:25
From: Eugene
Subject: Mathematics 6th grade

Here's the question: The GCF of my numerator and denominator is 5. The 
fraction is equivalent to 4/6. 

I tried listing all the multiples of 5 and none is divisible by 6 
except 30. But if I divide this by 6 it will be 5 and 5 x 4 is 20. But 
20/30's GCF is not 5, it's 10. HELP!

Date: 04/13/99 at 12:48:27
From: Doctor Peterson
Subject: Re: Mathematics 6th grade

Hi, Eugene.

Yes, this is a little tricky; I fell into almost the same trap you 
did. You're thinking well so far, but now you have to back up and ask 
why it didn't work. To let you practice avoiding the trap, I'll work a 
slightly different problem: the GCF will be 7 rather than 5, and the 
fraction will be equivalent to 3/6 rather than 4/6.

What you are doing is looking for some number N by which you can 
multiply the numerator and denominator, so that

    3xN    3
    --- = ---
    6xN    6

and the GCD of 3xN and 6xN is 7. It makes sense to try N = 7, as you 
did with 5; but the GCD of 21 and 42 is 21, not 7. What happened?

Let's factor the numerator and denominator of our new fraction:

    3xN    3xN
    --- = -----
    6xN   2x3xN

Do you see that the GCD will not be N, but 3xN, because 3 and 6 
already have a common factor, 3? In order to make the GCD be 7, then, 
N must be 7/3. Then

    3xN   7
    --- = --
    6xN   14

which is equivalent to 3/6, and the GCD of 7 and 14 is 7.

Were you surprised that N was not a whole number? You can multiply the 
numerator and denominator of a fraction by any fraction and it will 
still be equivalent; but the result will be two whole numbers only if, 
as in this case, there was a common factor (3).

So the reason our problem was tricky is that 3/6 is not in lowest 
terms, so equivalent fractions do not have to have whole-number 
multiples of the numerator and denominator. We can get to the answer 
very easily by first writing 3/6 in lowest terms as 1/2, then 
multiplying numerator and denominator by 7.

Now see what you can do with your problem. There are several ways to 
solve it, once you see what's wrong.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/