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 IntegerFirst 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*5^2Total 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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51% (01:40) correct
