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

Question

(1) x^2 is divisible by 9.

(2) x^4 is divisible by 9.

Bunuel · Accepted Answer

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.

gmatprav · Answer

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.

AndyNeedsGMAT · Answer

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

Bunuel · Answer

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  Divisibility/Multiples/Factors: Tips and hints
 .

Hope it helps.

Bunuel · Answer

5. Divisibility/Multiples/Factors 
     Theory   
 Divisibility and Remainders on the GMAT 
 Tips and hints: Divisibility/Multiples/Factors 
 3 Deadly Mistakes you must avoid in LCM-GCD Questions 
 How Well Do You Know Your Factors? 
 THE MOST EFFICIENT WAY TO STUDY LEAST COMMON MULTIPLES ON THE GMAT 
     Questions   
 The E-GMAT Primes Trio - 3 Ques on No. of factors & Prime factors 
 DS Questions 
 PS Questions

For other subjects:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

manojach87 · Answer

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 *   = 3^4 i.e. x=3
or 9*  = -3*-3*-3 *-3 = -3^4 i.e. x=-3
or 9*  = 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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If x is an integer, is x^3 divisible by 9?

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

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