Last visit was: 15 Jul 2025, 07:17 It is currently 15 Jul 2025, 07:17
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
gmatgambler
Joined: 17 Jun 2013
Last visit: 10 Jun 2015
Posts: 20
Own Kudos:
332
 [14]
Given Kudos: 20
Posts: 20
Kudos: 332
 [14]
1
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
741,566
 [7]
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,566
 [7]
4
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
User avatar
gmatprav
Joined: 25 Oct 2013
Last visit: 19 Nov 2015
Posts: 111
Own Kudos:
Given Kudos: 55
Posts: 111
Kudos: 180
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
AndyNeedsGMAT
Joined: 11 Oct 2012
Last visit: 26 Nov 2020
Posts: 28
Own Kudos:
Given Kudos: 74
GMAT 1: 610 Q42 V32
Products:
GMAT 1: 610 Q42 V32
Posts: 28
Kudos: 25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
russ9
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

Merging similar topics. Please refer to the discussion above.


Why cant we consider the case where x= sqrt 3; in that case x^3 will be divisible by 3sqrt 3
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,566
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ani781
Bunuel
russ9
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

Merging similar topics. Please refer to the discussion above.


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

Hope it helps.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Jul 2025
Posts: 102,576
Own Kudos:
741,566
 [3]
Given Kudos: 98,190
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,576
Kudos: 741,566
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
avatar
manojach87
Joined: 26 May 2016
Last visit: 18 Oct 2022
Posts: 13
Own Kudos:
Given Kudos: 5
GMAT 1: 640 Q49 V30
GMAT 1: 640 Q49 V30
Posts: 13
Kudos: 44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.


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
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,405
Own Kudos:
Posts: 37,405
Kudos: 1,013
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
102576 posts