Finding the LCM of Several NumbersDate: 11/06/2003 at 14:32:36 From: Roxana Subject: finding the lowest common denominator Do the numbers 65, 55, 25, and 35 have a common denominator? Multiplying them together, I get a bunch of different values. Date: 11/10/2003 at 13:04:26 From: Doctor Jason Subject: Re: finding the lowest common denominator Hi Roxana, Sometimes it's easier to see what's going on if we use smaller numbers. Suppose we want to find the least common multiple (LCM) of 4 and 6. How might we do that? We might start by finding the prime factors of each number: 4 = 2 * 2 6 = 2 * 3 Now, if we multiply 4 and 6, we get 24, which is certainly divisible by both numbers. But we'd like to see if there is a _smaller_ number that is also divisible by both 4 and 6. If we _do_ multiply 4 and 6, we get 4 * 6 = (2 * 2) * (2 * 3) = 2 * 2 * 2 * 3 Now, in order for this to be divisible by 4, we need to have a pair of 2's: 4 * 6 = 2 * 2 * 2 * 3 |___|_____________ 2*2 = 4 And in order for this to be divisible by 6, we need to have a 2 and a 3: 4 * 6 = 2 * 2 * 2 * 3 |___|_____________ 2*2 = 4 |___|_________ 2*3 = 6 So note that one of the 2's isn't serving any purpose: 4 * 6 = 2 * 2 * 2 * 3 ^ |___|_____________ 2*2 = 4 | | |___|_________ 2*3 = 6 | |_____________________ This one isn't contributing! Since it's not contributing, we can get rid of it: LCM(4,6) = 2 * 2 * 3 |___|_____________ 2*2 = 4 |___|_________ 2*3 = 6 = 12 If we remove either of the 2's, we get something that isn't divisible by 4. If we remove the 3, we get something that isn't divisible by 6. So there can't be any common multiple of 4 and 6 that is smaller than this. Does that make sense? Now, suppose we want to add a third number to the mix. That is, suppose we want to find the LCM of 4, 6, and 10? We already know that it can't be smaller than 12! So let's do this: LCM(4,6,10) = 2 * 2 * 3 * ? |___|_____________ 2*2 = 4 |___|_________ 2*3 = 6 What's missing? We can't make 10 by multiplying any combination of 2, 2, and 3. So we look at the prime factors of 10: 10 = 2 * 5 We've already got the two covered. All we need is the 5: LCM(4,6,10) = 2 * 2 * 3 * 5 |___|_____________ 2*2 = 4 |___|_________ 2*3 = 6 |_______|_____ 2*5 = 10 = 60 Can you see how we could extend this to find, say, the LCM of 4, 6, 10, and 21? Or any collection of numbers? Take another shot at your problem, and let me know what you come up with, okay? - Doctor Jason, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/