1.find the lcm of the list number is 9,6,21,14?

2.find the lcm of the list number is 12,15,45?

3. find the lcm of the list number is 7,5,6,15?

1.find the lcm of the list number is 9,6,21,14?

2.find the lcm of the list number is 12,15,45?

3. find the lcm of the list number is 7,5,6,15?

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Atlanta, GA

2-22-14

LCM is really fun. Here is what you do:

1. 9, 6, 21, 14 first write out the prime factorization of each number:

9= 3x3 6=2x3 21 = 2x7 14 = 7x2

Now, do this: 9= 3^2; 6=2x3; 21= 2x7; 14=2x7

The only number that is a common base between 2 or more numbers is 3--- take the 3 which has the larger exponent which would be 3^2--- cross out the other 3- you will not use it

Now take the common base 2 and only use the one which has the largest exponent-- all of the 2s have just a 1 for the exponent, so you will use just one of those 2s and cross out the other two-- you won't use them;

Now we have two 7s--they are equal, so just take one of them, get rid of the other --- in numerical order, this leaves us with 2 x 3^2 x 7 = 2 x 9 x 7 = 126= LCM

Now that you know how to do it, you can do numbers 2 and 3 by yourself! good luck :)

Saugus, MA

2) 12, 15, and 45

Find the LCM of the numbers.

Write the prime factorization of each number.

12=2x2x3

15=3x5

45=3x3x5

To be a multiple of 12, the LCM must contain all of 12's prime factors.

LCM=2x2x3 *so far, we aren't done yet !*

To be a multiple of 15, the LCM must contain all of 15's prime factors. We need the 3 and the 5, but we take only those factors that we don't already have ! We have the 3 but not the 5 so we use the 5 but we do not use the 3 again because we want the
*least* common multiple.

LCM=2x2x3x5 *so far !*

To be a multiple of 45, the LCM must contain all of 45's prime factors. We need two 3's and a 5, but again we take only those factors that we don't already have !

We have the 5 but we have only one 3 so we take one of the 3's.

LCM=2x2x3x5x3.

LCM=2x2x3x3x5

LCM=180

As a side note, if you realized that 15 is a factor of 45, you could have taken 45 and two 2's from the 12 and multiply 45x2x2=180.(we don't need the 3 from 12 because there is a 3 in 45 !)

Woodland Hills, CA

9 = 2^{0} . 3^2 .7^{0}

6 = 2 . 3 . 7 ^{0}

14 = 2 . 3^{0} . 7

21 = 2^{0 .} 3 . 7

14 = 2 3^{0} . 7

LCM = 2 . 3^2 . 7 = 126

After writing every number as a product of its prime factor

LCM = product of common factors with highest exponents. Keeping consistency we grant factors with

0 exponents to the numbers that don't have the factor.

Like 9 = 2^{0} . 3 ^{2 }

6 = 2 . 3

LCM = 3^{2} . 2 = 18

Greenville, SC

There are several methods for finding the LCM of two numbers, but the easiest for three or more is to use the factor tree method. For example, to find the LCM 9, 6, 21, and 14, use a factor tree for each number as shown.

9 6 21 14

3 3 2 3 3 7 2 7

Try to make sure your factors are in order from least to greatest. If any of the numbers have a common prime factor, write it
**once** and cross it off those numbers once. Write it down.

9 6 21 14

3 3
2 3 3 7
2 7

I am using a tablet, so I put my common factors in orange.

Write them down: 2

Continue to do this (next you would cross out three 3's, under the 9, 6, and 21) until no numbers have common factors. You should be left with a
3 under the 9, and a list of these common prime factors: 2, 3, 7. Write the remaining factors in your list once each. Multiply your numbers together, and you have the smallest number that can be evenly divided by each of your given numbers.

Answer: 2 x 3 x 7 x 3 = 126 (Check it out: 2 x 7 = 14, 3 x 3 = 9, and 14 x 9 = 126, so it can be divided by 14. You can check the other numbers, too.

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