If x is an integer that has exactly three positive divisors (these include 1 and x), how many positive divisors does x^3 have?
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?
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?
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
itself. So, x
can be 4, 9, 25, ... For example divisors of 4 are: 1, 2=prime, and 4 itself.
, 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
--> # 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
, and c
are prime factors of n
, and r
are their powers.
The number of factors of n
will be expressed by the formula (p+1)(q+1)(r+1)
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
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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