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.

Hi Bunuel, is it okay to reason stmt2 as below?

\(x^4\) is divisible by 9.
then either
a. x is divisible by 9 or
b. \(x^3\) is divisible by 9.

If a is true then b is also true.
if a is not true then b has to be true. therefore for \(x^4\) to be divisible by 9, \(x^3\) MUST be divisible by 9.
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Because we are told that x is an integer, while \(\sqrt{3}\) is not. Also, what does divisibility by \(3\sqrt{3}\) has to do with the question? We need to find whether x^3 divisible by 9.

Also, note two things:
1. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only).

2. On the GMAT when we are told that \(a\) is divisible by \(b\) (or which is the same: "\(a\) is multiple of \(b\)", or "\(b\) is a factor of \(a\)"), we can say that:
(i) \(a\) is an integer;
(ii) \(b\) is an integer;
(iii) \(\frac{a}{b}=integer\).

For more check [url=http://gmatclub.com/forum/divisibility-multiples-factors-tips-and-hints-174998.html]Divisibility/Multiples/Factors: Tips and hints
[/url].

Hope it helps.
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Answer: D

from 1, x^2 = 9n
multiplying both sides by x, we get
x*x^2 = x*9n
or x^3 = 9xn
let xn = p
then x^3 = 9p
i.e. x^3 is divisible by 9

Hence 1 is sufficient

from 2,
x^4 = 9m
now, since 9m should be something to the power of 4, there are only limited values of m

ex : 9 * [3*3] = 3^4 i.e. x=3
or 9*[ -3 * -3 ] = -3*-3*-3 *-3 = -3^4 i.e. x=-3
or 9*[9*81] = 3*3 * 3*3 *9*9 = 3*3*3*3*3*3*3*3 = 3^8 = (3^2)^4 i.e x = 9

in each case is either 3, -3 or a power of either of the two. hence 2 is also sufficient

and the answer is D