Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Sep 2014, 08:09

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If x, y and k are integers, is xy divisible by 3?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 11 Sep 2009
Posts: 129
Followers: 4

Kudos [?]: 172 [0], given: 6

GMAT Tests User
Re: If x, y and k are integers, is xy divisible by 3? [#permalink] New post 20 Sep 2009, 18:55
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
I believe the correct answer is D.

Basically, in order for xy to be divisible by 3, x or y HAVE TO be divisible by 3 on their own, since 3 is a prime number.

Statement 1: y = 2^(16) - 1:
The key here is knowing that 2^(odd integer) = 3n + 2, and 2^(even integer) = 3n + 1 (where n is a non-negative integer). Since 16 is an even integer, 2^16 will be equal to 3n + 1. Therefore 2^16 - 1 = 3n + 1 - 1 = 3n. Therefore y is divisible by 3, and Statement 1 is sufficient.

The reason that this rule works for 2^n when considering divisibility by 3 is as follows (sorry if this is confusing): if you take a number in the form of 3n + 1, and multiply it by 2, you get 6n + 2, which can be rewritten in the form 2*3n + 2 = 3m + 2. Multiplied again by 2, you get 6m + 4, which can be rewritten in the form 6(m+1) + 1 = 3p + 1. The number 2 (being 2^1) is in the form 3n + 2. Subsequently all numbers in the form 2^(odd number) can be written as 3n + 2, and all numbers in the form 2^(even number) can be written as 3n + 1. I remember learning this somewhere years ago, there might be a better explanation out there on the internet.

Statement 2: The sum of the digits of x equals 6^k :
For x to be divisible by 3, the sum of its digits must also be divisible by 3. It is given that the sum of the digits of x equals 6^k.

6^k = 3^k*2^k and therefore x is definitely divisible by 3. Therefore Statement 2 is also sufficient.

Since both Statement 1 and 2 are both sufficient by themselves, the correct answer is D.
Director
Director
User avatar
Joined: 01 Apr 2008
Posts: 909
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 15

Kudos [?]: 216 [0], given: 18

GMAT Tests User
Re: If x, y and k are integers, is xy divisible by 3? [#permalink] New post 20 Sep 2009, 20:51
Perfect! :)
Manager
Manager
User avatar
Joined: 25 Mar 2009
Posts: 57
Followers: 1

Kudos [?]: 10 [0], given: 9

Re: If x, y and k are integers, is xy divisible by 3? [#permalink] New post 21 Sep 2009, 08:00
What if k=0, then Sum digits of x =1, insuff

OA: A
Re: If x, y and k are integers, is xy divisible by 3?   [#permalink] 21 Sep 2009, 08:00
    Similar topics Author Replies Last post
Similar
Topics:
1 Experts publish their posts in the topic If x, y and k are integers, is xy divisible by 3? (1) y = shrive555 7 24 Oct 2010, 11:34
1 If x and y are integers, is xy divisible by 3 ? 1. (x+y)^2 bigfernhead 9 27 Apr 2009, 14:51
1 If x and y are integers, is xy + 1 divisible by 3? 1) when x rao 4 01 Aug 2008, 16:17
if x and y are integers is xy + 1 divisible by 3 ? 1. when x winsome 3 17 Mar 2007, 20:20
If X,Y r integers, is X*Y divisible by 3 ? 1) (X+Y)^2 allabout 7 08 Jan 2006, 12:06
Display posts from previous: Sort by

If x, y and k are integers, is xy divisible by 3?

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.