# Lesson: Prime Factorization

## Comment on Prime Factorization

### Are "distinct" prime factors

Are "distinct" prime factors defined as prime factors <10? (looking at some of the reinforcement activity links)

### "distinct" ≈ "unique"

"distinct" ≈ "unique"

Example #1: Consider the prime factorization of 88.
88 = (2)(2)(2)(11)
So, the distinct prime factors of 88 are 2 and 11

Example #2:
3042 = (2)(3)(3)(13)(13)
So, the distinct prime factors of 3042 are 2, 3 and 13

Does that help?

Cheers,
Brent

### Yes, that does help! I think

Yes, that does help! I think it was just a coincidence that the questions I saw had distinct prime factors <10. Thank you!!

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/s-is-the-set-of-all-integer-multiples-of-2486.html

I'm having trouble understanding the meaning of the sentence of Quantity A. Can you explain to me? Thanks.

Quantity A: The number of elements in the intersection sets S and T

The intersection of two sets refers to the values that the sets have in common.

For example: set X = {1, 2, 3, 4, 5, 6, 7} and set Y = {2, 4, 6, 8, 10, 12}

The intersection of the sets = 2, 4, 6, because these are the values that both sets have in common.

Does that help?

Cheers,
Brent

### 7*5=35 is also a divisor of

7*5=35 is also a divisor of 1540 I'm sure you chose to omit it

### You're right; 35 is also a

You're right; 35 is also a divisor of 1540
My intention wasn't to list all of the divisors.

Cheers,
Brent

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/if-x-2-7-4-2-7-0-what-is-the-largest-10612.html

I am missing something here, but not sure what. Could you explain to me why we minus one from the powers after factoring?

Factor 2^70 from right side to get: x = 2^70(2^4 - 1)

This is no different from factoring x^74 - x^70
We can factor out x^70 to get: x^74 - x^70 = x^70(x^4 - 1)
Notice that, if we take x^70(x^4 - 1) and expand it, we get back to x^74 - x^70

This kind of factoring is the same as 3x + 3 = 3(x + 1)
Once again, if we expand 3(x + 1), we get back to 3x + 3

It might help to review the lesson on GCF factoring: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...

Cheers,
Brent

### Please direct me to the

Please direct me to the module of where you taught how to write the Prime factorisation of large numbers

### The video at the top of the

The video at the top of the page shows how to find the prime factorization of all numbers.

Cheers,
Brent

### I have a question: how many

I have a question: how many different positive integer factors does 64*81*125 have? this is from kaplan test

### The strategy for this kind of

The strategy for this kind of question is covered here: https://www.greenlighttestprep.com/module/gre-integer-properties/video/828

In the meantime, here's my solution:

For the strategy covered in the video to work, we need only find the prime factorization of the given expression.
(64)(81)(125) = (2)(2)(2)(2)(2)(2)(3)(3)(3)(3)(5)(5)(5)
= (2^6)(3^4)(5^3)

So, the total number of positive factors = (6+1)(4+1)(3+1) = (7)(5)(4) = 140

Cheers,
Brent

### Hi Brent,

Hi Brent,

In this questio : https://gre.myprepclub.com/forum/what-is-the-smallest-positive-integer-that-is-non-prime-10760.html

9!=9*8*7*6*5*4*3*2*1
362880=5*2^7*3^^4*7

so the smallest integer may be evne 11 but ehy didnt you consider 11?

Thanks

You're correct to say that 9! = 362,880
However the prime factorization is incorrect.
We know this because 9! = 9 x 8 x 7 x 6 x 5 x 4 x 2 x 1
= (3 x 3) x (2 x 2 x 2) x 7 x (2 x 3) x 5 x (2 x 2) x 2 x 1
= (2⁷)(3³)(5)(7)

Once we know the prime factorization we can see that:
1, 2, 3, 4, 5, 6, 7, 8, and 9 are all factors of 9!
Also, 10 is a factor of 9! since 10 = (5)(2), and we can see one 5 and one 2 hiding in the prime factorization of 9!
Also, 12 is a factor of 9! since 12 = (2)(2)(3), and we can see two 2's and one 3 hiding in the prime factorization of 9!
14 is a factor of 9! since 12 = (2)(7), and we can see one 2's and one 7 hiding in the prime factorization of 9!
16 is a factor of 9! since 16 = (2)(2)(2)(2), and we can see four 2's hiding in the prime factorization of 9!
18 is a factor of 9! since 18 = (2)(3)(3), and we can see one 2 and two 3's hiding in the prime factorization of 9!
20 is a factor of 9! since 12 = (2)(2)(5), and we can see two 2's and one 5 hiding in the prime factorization of 9!
22 is NOT a factor of 9! since 22 = (2)(11), and we there are NO 11's hiding in the prime factorization of 9!

Does that help?

Cheers,
Brent

### why are we checking only for

why are we checking only for some numbers as 22 it means there must be atleast one prime inorder to check with the product so you have not choosen 11
eg : for 18- we have 2's and 3's in as prime factors
but for 11 - we doesnt have 11 or any number such that by multiplying with some x could yeild 18

is this correct

### The question asked us to find

The question asked us to find the smallest positive integer that is NON-PRIME and NOT a factor of 9!

Since there are zero 11's hiding in the prime factorization of 9!, we know that 11 is NOT a factor of 9!
However, we're looking for a NON-PRIME that isn't a factor of 9!

So, once I showed that 1, 2, 3, 4, 5, 6, 7, 8, and 9 ARE factors of 9!, I started testing other NON-PRIME values.

So, the next number I chose to examine was 10.
Once I was able to show that 10 IS a factor of 9!, I then tested the NEXT NON-PRIME value (12)
Once I was able to show that 12 IS a factor of 9!, I tested the next NON-PRIME value (14)
Once I was able to show that 14 IS a factor of 9!, I tested the next NON-PRIME value (16)
.
.
.
And so on, until I tested 22

Does that help?

### yes, that non prime wrord

yes, that non prime wrord makes a lot of difference .
Thanks

Why did you multiply (999)(9999) and how did you figure out it would be a multiple of 999 and 9999?

That's a very tough question (31% success rate)!

RULE: If j and k are integers, then jk is a multiple of j, and jk is is a multiple of k.
For example, since 35 = (5)(7), we know about 35 is a multiple of 5, and 35 is a multiple of 7.

Likewise, we know that (999)(9999) is a multiple of 999, and it's a multiple of 9999.

Does that help?

### In this video, is there any

In this video, is there any reason the prime factorization theorem does not say "any number greater than 1 is a prime or product of primes".

### But that's exactly what I say

Ha! I never noticed that before.
Yes, it should say 1 (not 2)

### thanks for clarifying Brent!!

thanks for clarifying Brent!! you are marvelous as usual.

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-the-product-of-6-distinct-positive-integers-is-6-6-what-is-the-min-321660.html

How would we approach this one as well?? Looks preety confusing, tried to backsolve it

This question doesn't represent what you'll see on test day.
The main issue is where the question states that all six integers must be different. This requires a lot of brute force to minimize the average.

### https://gre.myprepclub.com/forum

https://gre.myprepclub.com/forum/n-is-an-even-in-teger-17382.html
why is does not answer D because already say n is even integer zero is also even integer behalf that answer C should be.