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Intern  Joined: 18 Feb 2010
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If x is an integer that has exactly three positive divisors  [#permalink]

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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?

Thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 58445
Re: If x is an integer that has exactly three positive divisors  [#permalink]

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2
3
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?

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).

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.

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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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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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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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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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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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?

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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