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### Great Explanation of How to Find the Least Common Multiple

```Date: 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!!!
```
Associated Topics:
Middle School Factoring Numbers

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