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If the integer n has exactly three positive divisors, includ

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If the integer n has exactly three positive divisors, includ [#permalink] New post 18 Jan 2013, 08:43
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If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(A) 4
(B) 5
(C) 6
(D) 8
(E) 9

OG 11 #241.

Would someone mind explaining? I'm not satisfied with the explanation in the OG.
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Re: If the integer n has exactly three positive divisors, includ [#permalink] New post 18 Jan 2013, 09:09
Basically, the description says that this is the square of a prime number. So if you square that number, you will have a prime number raised to the fourth power.

That will have 5 factors. For a more detailed description, we have a free factors and multiples lesson on our site.
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Re: If the integer n has exactly three positive divisors, includ [#permalink] New post 18 Jan 2013, 09:47
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GMATBeast wrote:
If the integer n has exactly three positive divisors, including 1 and n, how many positive divisors does n^2 have?

(A) 4
(B) 5
(C) 6
(D) 8
(E) 9

OG 11 #241.

Would someone mind explaining? I'm not satisfied with the explanation in the OG.


Important property: the number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square. (A perfect square, is an integer that can be written as the square of some other integer. For example 16=4^2, is a perfect square).

Hence, since given that n has 3 (odd) divisors then n is a perfect square, specifically square of a prime. The divisors of \(n\) are: \(1\), \(\sqrt{n}=prime\) and \(n\) itself. So, \(n\) can be 4, 9, 25, ... For example divisors of 4 are: 1, 2=prime, and 4 itself.

Now, \(n^2=(\sqrt{n})^4=prime^4\), so it has 4+1=5 factors (check below for that formula).

Answer: B.

Else you can just plug some possible values for \(n\): say \(n=4\) then \(n^2=16=2^4\) --> # of factors of 2^4 is 4+1=5.

Answer: B.

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

So, the # of factors of x=a^2*b^3, where a and b are different prime numbers is (2+1)(3+1)=12.

Hope it's clear.
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Re: If the integer n has exactly three positive divisors, includ [#permalink] New post 19 Jan 2013, 09:25
quite simple..
take the example of 4...
it has 3 positive divisors (1,2,4)

Now, take the example of 16...
it has only 5 divisors..
so B is the ans
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Re: If the integer n has exactly three positive divisors, includ [#permalink] New post 30 Jan 2013, 01:02
gmat dose not requires us to remember much.

pick some numbers and see that the number must be a square of prime.

from this departure, we can infer B.

hard one
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Re: If the integer n has exactly three positive divisors, includ [#permalink] New post 03 Sep 2014, 05:50
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Re: If the integer n has exactly three positive divisors, includ   [#permalink] 03 Sep 2014, 05:50
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