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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]

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18 Jan 2013, 09:43
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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.
[Reveal] Spoiler: OA

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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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18 Jan 2013, 10:09
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]

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18 Jan 2013, 10:47
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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).

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.

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]

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19 Jan 2013, 10:25
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]

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30 Jan 2013, 02:02
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]

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03 Sep 2014, 06:50
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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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26 Nov 2015, 04:32
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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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01 Dec 2015, 15:42
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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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01 Mar 2017, 00:16
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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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08 Apr 2017, 07:56
It is 5. (b). If you try and experiment it, any number with three factors total, you will get 4 as one possibility as it is the only possibility there is. so you know that integer n=4. So 2 to the power of 4 is 16. 16 has 5 factors including itself and 1!

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If the integer n has exactly three positive divisors, includ [#permalink]

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08 Apr 2017, 23:22
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

because n will always be the square of a prime,
positive divisors of n^2 will always be 1, √n, n, n√n, n^2
5
B

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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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18 Apr 2017, 16:14
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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

Since n has exactly 3 positive divisors we can conclude that n is a perfect square of a prime number. For instance, let’s consider the prime number 3. Notice that 3^2 = 9, and the factors of 9 are 1, 3, and 9.

Thus, if we let n = 9, then 9^2 = 81.

The factors of 81 are 1, 81, 9, 27, and 3. Thus, n^2 has 5 positive divisors.

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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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07 Sep 2017, 07:18
Hi Scott,

Thanks for the explanation but I am bit lost here. In your explaination n^2 has three factors i.e 9 ( 1,3,9) but the question says n has three factors i.e 3 but three has only two factors ( 1,3). Where am I going wrong? Can you also explain the same with one more example?

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Re: If the integer n has exactly three positive divisors, includ [#permalink]

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07 Sep 2017, 07:25
santro789 wrote:
Hi Scott,

Thanks for the explanation but I am bit lost here. In your explaination n^2 has three factors i.e 9 ( 1,3,9) but the question says n has three factors i.e 3 but three has only two factors ( 1,3). Where am I going wrong? Can you also explain the same with one more example?

You should read a question and the solutions MUCH more carefully.

In Scott's example, n = 9 NOT n^2.

n = 9 has three factors: 1, 3, and 9. n^2 in this case will be n^2 = 9^2 = 3^4 and it will have 4 + 1 = 5 factors.
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Re: If the integer n has exactly three positive divisors, includ   [#permalink] 07 Sep 2017, 07:25
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