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If the integer n has exactly three positive divisors, includ

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

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New post 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).

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]

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

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

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New post 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.

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

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New post 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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New post 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]

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Re: If the integer n has exactly three positive divisors, includ   [#permalink] 17 Oct 2019, 06:14
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