Great Explanation of How to Find the Least Common MultipleDate: 10/24/2007 at 16:30:46 From: Dustin Subject: Time Difference If you set (3) alarm clocks to go off at (3) different times--when will they all ring at the same time? For example, set clock #1 to ring at 25 min. / clock #2 to ring at 30 min. / clock #3 to ring at 35 min. How many minutes will it take for them all to ring at the same time? I was told the answer of 1050 minutes, but I am having trouble figuring out the formula. Date: 10/24/2007 at 16:43:09 From: Doctor Ian Subject: Re: Time Difference Hi Dustin, The "formula" is really more of a concept. You want to find the smallest number that is a multiple of 25, 30, and 35, usually called the Least Common Multiple (or LCM) of the numbers. Multiplying the numbers together gives A common multiple, but not necessarily the LEAST one. Let's look at the prime factors: 25 = 5 * 5 30 = 2 * 3 * 5 35 = 5 * 7 If we want a number that is a multiple of all these, it would have to have the union of the common factors: 25 = 5 * 5 30 = 2 * 3 * 5 35 = 5 * 7 LCM = 2 * 3 * 5 * 5 * 7 = 1050 So 1050 is, in fact, the LCM. All three clocks will ring after 1050 minutes, and that is the first time it will happen. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 10/24/2007 at 17:21:28 From: Dustin Subject: Thank you (Time Difference) Wow - thank you for the quick and excellent answer. I am confused on one thing: I understand the factors 25 = 5*5 30 = 2*3*5 35 = 5*7 But why in the LCM formula, is only 2*3*5*5*7 used and not all the numbers? What determines the selected numbers? It seems to me that the formula would include all of the numbers. Thanks again. Date: 10/24/2007 at 21:34:35 From: Doctor Ian Subject: Re: Time Difference Hi Dustin, We want the smallest number that is divisible by 25, 30, and 35. To be divisible by We need these prime factors ------------------ --------------------------- 25 5, 5 30 2, 3, 5 35 5, 7 So imagine we have a little shopping basket, and we take it to the 'prime factors store'. We toss in a couple of 5's, to take care of 25: basket: 5, 5 <-- We can make 25 from these Next, we need to 'shop' for 30. But we already have the 5 we need for 30, right? All we have to add to our basket are 2 and 3: basket: 2, 3, 5, 5 <-- We can make 25 from these, and 30, too. Finally, we shop for 35. Again, we already have the 5 we need for this, so we just have to add a 7: basket: 2, 3, 5, 5, 7 <-- We can make 25 from these, and 30, too. And 35. If we took any one of these prime factors out of the basket, we would lose the ability to make at least one of the numbers. So this is the smallest possible set of prime factors, and their product is 1050. We could toss more prime factors in, and while we would get a common multiple, it wouldn't be the LEAST common multiple. And in fact, that's what happens if we just multiply the numbers--it's like shopping for each number independently, without taking into consideration what we've got from the others: ________________________ These we need. | | | | | 25 * 30 * 35: 2, 3, 5, 5, 5, 5, 7 |__|_________ These we don't. Does that make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 10/25/2007 at 15:53:30 From: Dustin Subject: Thank you (Time Difference) Thank you again, the shopping basket did it all for me. I understand why we would only use certain prime numbers in the equation. This site rocks!!! |
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