Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is based on a lesson at Khan Academy that I didn't understand.

In the lesson, the instructor uses the number 512 as an example and the entire prime factorization consists of three groups of three 2s. However, not every number has such a nice prime factorization and I don't understand how to use prime factorization to determine a cube root.

For example, if the number is 1000, I know the cube root is 10, but how do I get there using prime factorization.

The prime factorization of 1000 is 2 * 2 * 2 * 5 * 5 * 5. Since it's six prime numbers, I can't get 3 equal groups of the same number like I could with 512. In another forum someone said to use three groups of the smallest factor, but three 2s doesn't get me 10.

What am I missing?

share|improve this question
    
You're missing a copy of $2$. $1000 = 2^{3}5^{3}$. –  Isaac Solomon Apr 23 '13 at 19:16

3 Answers 3

up vote 4 down vote accepted

You're missing a third factor of $2$ in your prime factorization of $1000$:

$$1000 = 2\cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5 = (2^3\cdot 5^3) = (2\cdot 5)^3 = (2\times 5) \cdot (2\times 5) \cdot (2 \times 5) = 10 ^3$$

share|improve this answer
    
Can you see how to reorder the terms, to get three groups/factors of $10$? –  amWhy Apr 23 '13 at 19:23
    
We simply rearrange the order of the factors (which doesn't change their product): $2\cdot 2\cdot 2 \cdot 5 \cdot 5\cdot 5 = 2 \cdot 5 \cdot 2 \cdot 5 \cdot 2 \cdot 5 = 10 \cdot 10 \cdot 10$ –  amWhy Apr 23 '13 at 19:25
    
I think I get it. I was thinking I had to use the numbers in order. I forgot that the order is arbitrary. DOH! So, my three groups are all 2*5. Now that I realize this, I think I can apply it to other numbers. Thanks! –  user74014 Apr 23 '13 at 19:27
    
You're welcome! Yes, multiplication is commutative on the integers: so $m\times n = n\times m$, etc. –  amWhy Apr 23 '13 at 19:31
    
@amWhy: you do such a great job of following up! +1 –  Amzoti Apr 24 '13 at 0:35

The prime factorization for 1000 is 2*2*2*5*5*5. So you have three groups you're trying to find.

share|improve this answer
    
You're just missing another 2 in your prime factorization. So it looks like you're probably doing everything else correctly in finding the 3 groups of numbers, etc. –  agktmte Apr 23 '13 at 19:17
    
I corrected my error, but I don't get 10 with three 2s or with three 5s, so I still don't get it. –  user74014 Apr 23 '13 at 19:21
    
Combine the groups: 2*2*2*5*5*5 = 2*5 * 2*5 * 2*5 if you reorder them, so it becomes 10*10*10, just rearrange the prime factors. –  agktmte Apr 23 '13 at 19:22

What you're missing is there are indeed 3 factors of 2 in 1000. Remember, when you multiply a number by 10, you add a 0 to the end of it, so it should be easy to see that

$$1000=10\times10\times10$$

share|improve this answer
    
I corrected my error, but I don't get 10 from three 2s. I realize it's easy to memorize the cube root of a number like 1000. That was just an example. The question was how do I use prime factorization to get a cube root? –  user74014 Apr 23 '13 at 19:23
    
@Ghodmode It looks like amWhy has that part covered. –  Mike Apr 23 '13 at 19:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.