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Date: 12/31/97 at 17:14:50
From: Mike Newman
Subject: Denominators

I have a problem with different denominators... 4/5+5/6+7/16 = ? 
What is the best and fastest way to find the LCD?

Mike Newman

Date: 01/01/98 at 14:52:20
From: Doctor Mike
Subject: Re: Denominators


It's always nice to hear from another Mike - and there are a lot of us 
out there. 
The short answer to your question is "Prime Factorization," but let me 
explain that. Prime numbers are those whole numbers greater than 1 
which are divisible only by themselves and one. Your first denominator 
5 is a prime number, but the second is not because 6 = 2*3. Also, the 
third denominator is not because 16 = 2*2*2*2. These primes you 
multiply together to get the number are called the prime factors. 
Notice that some of the factors get repeated. 
IF you do not have to find the *least* common denominator, but just 
any old common denominator, then you multiply the three together, 
i.e., 5*6*16 = 480.  This is the easiest common denominator to find, 
but usually not the easiest one to work with. The *least* common 
denominator has to have all the factors for all the denominators.  
In this case it is
       2*2*2*2*3*5 = 240
It has the four 2's for 16. It has the 2 and 3 for 6. It has the 5 
for 5. But notice that one of the 2's for 16 also serves as the 2 for 
6. This is somewhat of an improvement over 480. 

Sometimes the improvement is much better. For instance, if the 
denominators are 60 and 10 and 15 and 3, the product of all 4 
denominators is 27000, but the least common denominator is 2*2*3*5 = 
60 .... quite a bit simpler! So, the best and fastest way to find the 
least (smallest) denominator that works is to factor all the 
denominators until you cannot factor them any more, and then construct 
the new denominator to have *just enough* of all the prime factors 

I hope this helps. 
Since you asked about finding the LCD I assume that you can take it 
from there to convert the 3 fractions to equivalent fractions with 240 
as the denominator, and then add.  

-Doctor Mike,  The Math Forum
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Associated Topics:
Elementary Fractions

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