If x is an integer that has exactly three positive divisors

Question

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

Bunuel · Accepted Answer

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.

anordinaryguy · Answer

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.

nades09 · Answer

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!

TGC · Answer

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

!

unverifiedvoracity · Answer

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.

Abhishek009 · Answer

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...

Mission2027 · Answer

Hi  Bunuel ,

Can you please elaborate the following:

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

Warm Regards,

Bunuel · Answer

First of all check below:

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.

Next, perfect squares, integers with positive even powers, have odd number of positive factors. For example, 12^4 = 2^8*3^4 has (8 + 1)(4 + 1) = 9*5 = 45 = odd number of factors. More generally, the number of factors of n^even, where  n^{even}=(a^p*b^q*c^r)^{even}=a^{p*even}*b^{q*even}*c^{r*even}=a^{even}*b^{even}*c^{even} , ( a ,  b , and  c  are prime factors of  n  and  p ,  q , and  r  are their powers.) will be (even + 1)(even + 1)(even + 1) = odd*odd*odd = odd

Now, if the number of factors of x is 3, then x must be not only a perfect square but a square of a prime number. For example: 
4 = 2^2 has (2 + 1) = 3 factors: 1, 2, and 4;
9 = 3^2 has (2 + 1) = 3 factors: 1, 3, and 9; 
25 = 5^2 has (2 + 1) = 3 factors: 1, 5, and 25; 
...

Hope it's clear.

MBAHOUSE · Answer

If x has exactly three positive divisors, it means x is a perfect square of a prime number. The reason for this is that a number with three positive divisors has to be of the form p^2, where p is a prime number.

Now, if x = p^2, then x^3 = (p^2)^3 = p^6. The prime factorization of p^6 is p * p * p * p * p * p. To find the number of positive divisors of x^3, we need to consider all possible combinations of these prime factors.

For each prime factor, we can choose to include it zero times, one time, two times, three times, four times, five times, or six times. Each choice will give us a distinct divisor of x^3. Therefore, the total number of positive divisors of x^3 is (6 + 1) = 7.

So, if x has exactly three positive divisors (which implies x = p^2, where p is a prime), then x^3 will have 7 positive divisors.

Paras96 · Answer

If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?

From the information available, we can infer that x = p^2 where p is any prime number

Now, x^3 = (p^2)^3 = p^6

Therefore no. of factors i.e. divisors of p = (6+1)=7

Hence D

sarthak1701 · Answer

Let's take x = 9 which has exactly 3 divisors (1,3 and 9)

9^3 = (3^2)^3 = 3^6

Factors of 3^6 = (6+1) = 7. Ans. D

SuprotimRoy · Answer

We have 2 odd number of factors in the options.

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

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