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# If positive integer X has 3 prime factors and 8 total factors, then..

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Intern
Joined: 25 Jul 2017
Posts: 14
If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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17 Sep 2017, 16:15
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25% (medium)

Question Stats:

74% (00:44) correct 26% (23:23) wrong based on 31 sessions

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If positive integer X has 3 prime factors and 8 total factors, then how many total factors will $$x^{n}$$ have where n is a positive integer?

A. 3
B. 5
C. $$n^{3}$$
D. $$(n+1)^{3}$$
E. n^6

I am struggling with primes and factors. Any and all advice is appreciated.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

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Math Expert
Joined: 02 Aug 2009
Posts: 5952
Re: If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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17 Sep 2017, 17:35
bkastan wrote:
If positive integer X has 3 prime factors and 8 total factors, then how many total factors will $$x^{n}$$ have where n is a positive integer?

A. 3
B. 5
C. $$n^{3}$$
D. $$(n+1)^{3}$$

I am struggling with primes and factors. Any and all advice is appreciated.

Hi...

Since there are 3 prime factors, let them be a,b,c...
Now since total factare 8, and 8 =2*2*2=(1+1)(1+1)(1+1) only possiblity is that each of the prime factor is used once..
That is x=a*b*c...
So $$x^n=a^n*b^n*c^n$$..
Total number of factors=(n+1)(n+1)(n+1)=(n+1)^3
D
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Intern
Joined: 25 Jul 2017
Posts: 14
Re: If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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18 Sep 2017, 03:39
chetan2u thank you for the help. Would you mind explaining the relevance of breaking down 8 into 2*2*2 and how this indicates that each factor is only used once? I'm still a tad lost.
Veritas Prep GMAT Instructor
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Location: Pune, India
Re: If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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18 Sep 2017, 04:47
1
1
bkastan wrote:
If positive integer X has 3 prime factors and 8 total factors, then how many total factors will $$x^{n}$$ have where n is a positive integer?

A. 3
B. 5
C. $$n^{3}$$
D. $$(n+1)^{3}$$
E. n^6

I am struggling with primes and factors. Any and all advice is appreciated.

Suggest you to check out this post first: https://www.veritasprep.com/blog/2010/1 ... ly-number/

It shows you why and how prime factorisation is done for a number N.

If a, b and c are the only prime factors of N, we can write N as

$$N = a^x * b^y * c^z$$

and the total number of factors in this case will be $$(x+1)*(y+1)*(z+1)$$

(all explained in the post)

Coming to our question,

Since X has 3 prime factors, say a, b and c and 8 total factors,

$$8 = (x+1)*(y+1)*(z+1)$$
8 is the product of three integers, each greater than 1.
So 8 = 2*2*2
x =1 , y= 1, z = 1

$$X = xyz$$
$$X^n = (xyz)^n = x^n * y^n * z^n$$

Total number of factors of $$X^n = (n+1)*(n+1)*(n+1) = (n+1)^3$$

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Joined: 13 Mar 2017
Posts: 610
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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19 Sep 2017, 02:32
bkastan wrote:
If positive integer X has 3 prime factors and 8 total factors, then how many total factors will $$x^{n}$$ have where n is a positive integer?

A. 3
B. 5
C. $$n^{3}$$
D. $$(n+1)^{3}$$
E. n^6

I am struggling with primes and factors. Any and all advice is appreciated.

8 = 2*2*2
That means the prime numbers have powers as 1 , 1,1
So lets say the prime numbers be a,b,c
So, X = a^1* b^1 * c^1

and X^n = a^n *b^n *c^n

and hence total factors of this number = (n+1)(n+1)(n+1) = (n+1)^3

--== Message from GMAT Club Team ==--

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If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: If positive integer X has 3 prime factors and 8 total factors, then..   [#permalink] 19 Sep 2017, 02:32
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