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If x is an integer that has exactly three positive divisors [#permalink]
16 May 2012, 19:10

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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?

Re: Number of positive divisiors of X^3 [#permalink]
16 May 2012, 20:33

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.

Re: If x is an integer that has exactly three positive divisors [#permalink]
17 May 2012, 00:10

1

This post received KUDOS

Expert's post

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.

Re: If x is an integer that has exactly three positive divisors [#permalink]
17 May 2012, 12:56

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.

Re: If x is an integer that has exactly three positive divisors [#permalink]
30 Sep 2013, 03:11

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Re: If x is an integer that has exactly three positive divisors [#permalink]
30 Sep 2013, 10:52

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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Re: If x is an integer that has exactly three positive divisors [#permalink]
13 Dec 2014, 10:33

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