LCM

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Psychguy98

Junior Member
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Dec 17, 2010
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What is the least common multiple of the following numbers? 15, 39, 30, 21, 70

a. 2,370
b. 3, 270
c. 7,320
d. 2,730

I really don't know where to begin? 15 = 1 , 3, 15 ?
 
First find the prime factorization of each number:

\(\displaystyle 15=3\cdot 5\)
\(\displaystyle 39=3\cdot 13\)

See if you can do the rest of them.

Then multiply together the highest power of each prime that appears.
So, for example if we were just computing the lcm of 15 and 39, then the answer would be \(\displaystyle 3\cdot 5\cdot 13=195\)

If you get stuck, show me where you're stuck and I'll help you through the rest of it.
 
so for 30 = 3 and 10, 2 and 15

21 = 3 and 7

70 = 7 and 10, 2 and 35, 14 and 5 . Like that
 
Psychguy98 said:
so for 30 = 3 and 10, 2 and 15

21 = 3 and 7

70 = 7 and 10, 2 and 35, 14 and 5 . Like that
Click to expand...

Not quite "like that."

In a prime factorization, all of the factors are prime numbers.

For instance, we can write 12 as 3*4, but that is not a prime factorization of 12. We can write 12 as 2* 6, but that's not a prime factorization either.

We might START by writing 12 as 3*4, but then we have to look at 4 and say "Oh wait! 4 is not a prime number. I can break 4 down into 2*2."

So, 12 = 3*4
12 = 3*2*2
Look again....are all of our factors prime numbers? Yes, they are. So the prime factorization of 12 is 3*2*2, or using exponents, 3*2[sup:3deb0c6g]2[/sup:3deb0c6g].

Now work on the prime factorizations of 30 and 70.
 
for 30 we have 3 and 10. 3 is a prime number. 10 can be broken down into 2 and 5. 2 and 5 are prime numbers. 3 * 2 *5

70 we have 7 and 10. 7 is prime. 10 can be broken down into 2 and 5. Both are prime. 7 *5 * 2
 
Psychguy98 said:
for 30 we have 3 and 10. 3 is a prime number. 10 can be broken down into 2 and 5. 2 and 5 are prime numbers. 3 * 2 *5

70 we have 7 and 10. 7 is prime. 10 can be broken down into 2 and 5. Both are prime. 7 *5 * 2
Click to expand...

Yes! You've found those prime factorizations correctly. Now, go back and look at DrSteve's directions for how to find the Least Common Multiple.

I'll give you a completed example of a DIFFERENT problem. Find the least common multiple of 8, 24, 60 and 75

Write the prime factorizations for each of the numbers (using exponents if there are multiple instances of a prime factor):

8 = 2[sup:282sm9ls]3[/sup:282sm9ls]
24 = 2[sup:282sm9ls]3[/sup:282sm9ls]*3
60 = 2[sup:282sm9ls]2[/sup:282sm9ls]*3*5
75 = 3*5[sup:282sm9ls]2[/sup:282sm9ls]

Now, write down a product using each of the DIFFERENT prime factors you see, using the highest power that appears on each of those primes:

2[sup:282sm9ls]3[/sup:282sm9ls]*3*5[sup:282sm9ls]2[/sup:282sm9ls]

That product represents the LCM of the numbers 8, 24, 60 and 75. You can do the multiplication to get 600 as the LCM.

Use the same approach on your problem.
 
15 = 5 and 3

39 = 3 and 13

30 = 2 and 15 ( 5 and 3)

21 = 3 and 7

70 = 7 and 10 ( 5 and 2)

so 5^3 * 3^4 * 13 * 2^2 * 7^2 ?
 
OK. You're getting there. But your last step isn't correct.

Let me rewrite your prime factorizations a bit nicer first:

\(\displaystyle 15 = 3\cdot 5\)
\(\displaystyle 39 = 3\cdot 13\)
\(\displaystyle 30 = 2\cdot 3\cdot 5\)
\(\displaystyle 21 = 3\cdot 7\)
\(\displaystyle 70 = 2\cdot 5\cdot 7\)

Now you wrote that the lcm is \(\displaystyle 5^3 \cdot 3^4 \cdot 13 \cdot 2^2 \cdot 7^2\). This is incorrect. What you have written down is simply the product of the given numbers, ie. this is the same as just multiplying all the original numbers together. Do you see why?

Although the product is a common multiple of all the given numbers, it is usually not the least common multiple. To get the least common multiple you only take each prime factor that appears once, and you take the highest power of that prime that appears in any of the factorizations. For example, the highest power of \(\displaystyle 2\) that appears is simply \(\displaystyle 2\). So you would use \(\displaystyle 2\) in the lcm, not \(\displaystyle 2^2\). See if you can write down the answer now.


Side note: When writing down prime factorizations it is best to write the primes in increasing order. There is a good theoretical reason for this, but most importantly for you it will be more readable and you will be less likely to make careless errors. So for example notice that I write

\(\displaystyle 15 = 3\cdot 5\) as opposed to \(\displaystyle 15 = 5\cdot 3\)

The first of these is known as canonical form. The word "canonical" is just a sophisticated synonym for "natural"
 
15 = 3,5

39 = 3 and 13

30 = 2, 3 and 5

21 = 3 and 7

70 = 2, 5, 7


so 2 * 3 * 5 * 7 * 13
 
Yes - looks like you got it.

Notationwise, if you're not using latex it's better to use * rather than comma:

15=3*5, etc.
 
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