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

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

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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.
[Reveal] Spoiler: OA

Kudos [?]: 2 [0], given: 8

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

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New post 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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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New post 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.

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

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New post 18 Sep 2017, 04:47
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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 \(X^n = (n+1)*(n+1)*(n+1) = (n+1)^3\)

Answer (D)
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Re: If positive integer X has 3 prime factors and 8 total factors, then.. [#permalink]

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New post 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

Answer D
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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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