If x is an integer, is x^3 divisible by 9?
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
(1) x^2 is divisible by 9.
(2) x^4 is divisible by 9.
Statement 1: x² is divisible by 9.
x³ = x² * x
So if x² is divisible by 9, then x³ is divisible by 9.
Sufficient.
Statement 2: x� is divisible by 9.
9 = 3 * 3
So if x� is divisible by 9, then among the prime factors of x� are at least two 3's.
From the question we know that x is an integer. Meanwhile, 3 is prime.
If x� is divisible by 9, then x� = 3 * 3 * (some other integer). To get an integer x, the 4th root of the right side of the equation has to be an integer.
In order for the 4th root of the right side to be an integer, we need four sets of the same prime factors. So we need a total of four 3's among the prime factors of x�.
So, since x is an integer and 3 is prime, if there are among the prime factors of x� two 3's, there must be four 3's.
So x� = 3� * (some other integer)�.
Therefore, x³ = 3³ * (some other integer)³.
3³ is divisible by 9. So x³ is divisible by 9.
Sufficient.
The correct answer is D.
