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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
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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.
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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?