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If x is an integer that has exactly three positive divisors

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If x is an integer that has exactly three positive divisors  [#permalink]

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New post 16 May 2012, 20:10
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If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?

A. 4
B. 5
C. 6
D. 7
E. 8

I am not sure how the answer is derived here
If x=2, then X^3 =8 and 8 has 4 divisors - 1,2,4,8
But if x=9, then 9^3 =3^6, will have 7 divisors. So isn't the number of positive divisors dependent on the value of x?

Please explain

Thanks
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Re: If x is an integer that has exactly three positive divisors  [#permalink]

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New post 17 May 2012, 01:10
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nades09 wrote:
If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?

A. 4
B. 5
C. 6
D. 7
E. 8

I am not sure how the answer is derived here
If x=2, then X^3 =8 and 8 has 4 divisors - 1,2,4,8
But if x=9, then 9^3 =3^6, will have 7 divisors. So isn't the number of positive divisors dependent on the value of x?

Please explain

Thanks


x cannot be 2, because 2 has only two divisors 1 and 2, not three as given in the stem.

If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?
A. 4
B. 5
C. 6
D. 7
E. 8

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

Now, \(x^3=(\sqrt{x})^6=prime^6\), so it has 6+1=7 factors (check below for that formula).

Answer: D.

Else you can just plug some possible values for \(x\): say \(x=4\) then \(x^3=64=2^6\) --> # of factors of 2^6 is 6+1=7.

Answer: D.

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: Number of positive divisiors of X^3  [#permalink]

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New post 16 May 2012, 21:33
If x is an integer that has exactly three positive divisors (1, x, y)
=> It means x is a square number with x = y^2
So possible divisors of X^3 could be :
1, y, y^2 ---- (Divisors of x)
y^3, y^4 ---- (Additional Divisors of x^2)
y^5, y^6 ---- (Additional Divisors of x^3)

Regarding nbr 2 => It has only 2 divisors and those are 1 and 2.

Hope it clarifies situation.
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Re: If x is an integer that has exactly three positive divisors  [#permalink]

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New post 17 May 2012, 13:56
Thanks very much both of you for the responses.
Although I could get that x is a perfect square, since it has 3 factors including itself, I think I failed to correlate x = y^2.
Once that is done, I think x= y^6 and the number of positive divisors will be 7.

Thanks again for the explanations!
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Re: If x is an integer that has exactly three positive divisors  [#permalink]

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New post 30 Sep 2013, 11:52
Hit and trial one such number we have '4'

Since factors of '4'are {1,2,4}

4^3=64

Number of factors/divisors of 64= 2^6

We know that when a number is expressed as a product of the prime factors as below:

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

Then no. of divisors = (x+1)*(y+1)*(z+1)

Then here (6+1) = 7

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If x is an integer that has exactly three positive divisors  [#permalink]

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New post 04 Jun 2016, 20:54
X has odd number of factors, so x is a perfect square. Once you know that x is a perfect square, you know that \(x^3\) is also a perfect square. Perfect squares have odd number of factors and in the options, only Option D is odd. So, answer is D.
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Re: If x is an integer that has exactly three positive divisors  [#permalink]

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New post 05 Jun 2016, 00:19
nades09 wrote:
If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?

A. 4
B. 5
C. 6
D. 7
E. 8

I am not sure how the answer is derived here
If x=2, then X^3 =8 and 8 has 4 divisors - 1,2,4,8
But if x=9, then 9^3 =3^6, will have 7 divisors. So isn't the number of positive divisors dependent on the value of x?

Please explain

Thanks


Consider simple numbers like 4

4 has 3 divisors, 1 , 2 and 4

Now 4^3 = 64

Factors of 64 are 1, 2, 4, 8, 16, 32, 64

So, THere are 7 factors of x^3

Hope this helps...

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Re: If x is an integer that has exactly three positive divisors  [#permalink]

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Re: If x is an integer that has exactly three positive divisors   [#permalink] 18 Feb 2019, 01:28
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