If x, y and k are integers, is xy divisible by 3? : DS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 08:29

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 341 [0], given: 6

Re: If x, y and k are integers, is xy divisible by 3? [#permalink]

### Show Tags

20 Sep 2009, 18:55
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Joined: 01 Apr 2008
Posts: 897
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 28

Kudos [?]: 645 [0], given: 18

Re: If x, y and k are integers, is xy divisible by 3? [#permalink]

### Show Tags

20 Sep 2009, 20:51
Perfect!
Manager
Joined: 25 Mar 2009
Posts: 55
Followers: 1

Kudos [?]: 17 [0], given: 9

Re: If x, y and k are integers, is xy divisible by 3? [#permalink]

### Show Tags

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
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.