If x is an integer, is x^3 divisible by 9?

Important thing to know when solving this problem is that exponentiation does not "produce" primes. For example, for integer x if x^(positive integer) is divisible by 3, then x must be divisible by 3. How else would 3 appear in x^(positive integer)?

(1) x^2 is divisible by 9 --> x^2 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.

(2) x^4 is divisible by 9 --> the same here: x^4 is divisible by 3^2 --> x is divisible by 3 --> x^3 is divisible by 3^3*k, thus it's divisible by 9 too. Sufficient.

Answer: D.

Hope it's clear.
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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.

Why cant we consider the case where x= sqrt 3; in that case x^3 will be divisible by 3sqrt 3

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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5. Divisibility/Multiples/Factors

• Theory
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  Tips and hints: Divisibility/Multiples/Factors
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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

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