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## Comment on

Prime Factorization## Are "distinct" prime factors

## "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

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

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

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/s-is-the-set-of-all-integer-multiples-o...

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

## 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

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)

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-x-2-7-4-2-7-0-what-is-the-largest-10...

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

## 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

## 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,

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

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/what-is-the-smallest-positive-integer-t...

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

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

Thanks

## Really confused about this

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

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/s-is-the-set-of-all-integer-multiples-o...

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

## But that's exactly what I say

Ha! I never noticed that before.

Yes, it should say 1 (not 2)

## thanks for clarifying Brent!!

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

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

## Question link: https:/

Question link: https://gmatclub.com/forum/if-the-product-of-6-distinct-positive-integer...

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

why is does not answer D because already say n is even integer zero is also even integer behalf that answer C should be.

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/n-is-an-even-in-teger-17382.html

Sorry, I'm not sure what you're saying.

We're looking for DIFFERENT prime factors.

So, for example, if n = 6, the prime factors are 2 and 3 (2 DIFFERENT prime factors)

This means 2n = 12, the prime different factors of 12 are 2 and 3 (2 DIFFERENT prime factors)

Does that help?